Evaluate ∫ (2 x + 3)/(x^2 + 3 x + 5) dx.

(x (2 x + 3))/(x^2 + 3 x + 5) + C

Indefinite Integrals for CUET Maths

Q: What is an indefinite integral?

An indefinite integral of f is any function F whose derivative is f, called an antiderivative, written ∫ f(x) dx = F(x) + C. The constant C is there because adding a constant does not change a derivative, so two antiderivatives of the same function always differ by a constant.

Q: How does integration by substitution work?

Choose a new variable t for an inner expression whose derivative also appears in the integrand, up to a constant factor. Writing dt for that derivative times dx turns the integral into a standard form in t; integrate it, then put the inner expression back for t.

About 5 min to learn · 90 questions

Indefinite integrals are part of Section A of CUET Maths, the section every candidate answers, whether you take Mathematics or Applied Mathematics, and the next chapter's areas and the differential equations after it are built on them. Most questions are a known form in disguise, so the marks go to whoever spots the method fastest. Learn each method here, from the standard forms to substitution, trigonometric identities, partial fractions and integration by parts, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area: Section A1

On this page · 11 sections

01Integration as the reverse of differentiation
02Integration by substitution
03Integration with trigonometric identities
04Integrals of particular functions
05Integration by partial fractions
06Integration by parts
07How CUET asks this chapter
08Exam shortcuts
09Common mistakes
10Quick revision
11Check your understanding

Integration as the reverse of differentiation

Run the ball backwards. Suppose all we know is its velocity, $20 - 10t$ m/s, and we want its height. We need a function whose derivative is $20 - 10t$ : $20t - 5t^2$ works, and so does $20t - 5t^2 + 1.5$ , or any other constant added on. Finding such a function is integration.

What is an indefinite integral?

An indefinite integral of $f$ is any function $F$ whose derivative is $f$ , called an antiderivative, written $\int f(x)\,dx = F(x) + C$ . The constant $C$ is there because adding a constant does not change a derivative, so two antiderivatives of the same function always differ by a constant.

Every derivative you know gives an integral by reading it backwards.

Integral	Result	Integral	Result
$\int x^n\,dx$ , $n \ne -1$	$\frac{x^{n+1}}{n+1}$	$\int \frac{1}{x}\,dx$	$\log\lvert x \rvert$
$\int e^x\,dx$	$e^x$	$\int a^x\,dx$	$\frac{a^x}{\log a}$
$\int \sin x\,dx$	$-\cos x$	$\int \cos x\,dx$	$\sin x$
$\int \sec^2 x\,dx$	$\tan x$	$\int \operatorname{cosec}^2 x\,dx$	$-\cot x$
$\int \sec x\tan x\,dx$	$\sec x$	$\int \operatorname{cosec} x\cot x\,dx$	$-\operatorname{cosec} x$

Add $C$ to each. The integral of a sum is the sum of the integrals, and a constant factor comes outside: $\int kf\,dx = k\int f\,dx$ . There is no such rule for a product: $\int fg\,dx$ is not $\int f\,dx \cdot \int g\,dx$ .

Example 1 · Term by term

Evaluate $\int \left(x^3 - \frac{2}{x} + e^x\right)dx$ .

$\int x^3\,dx = \frac{x^4}{4}$ , $\int \frac{2}{x}\,dx = 2\log|x|$ and $\int e^x\,dx = e^x$ .
Combine them with one constant.

Answer

\frac{x^4}{4} - 2\log|x| + e^x + C

Example 2 · Fixing the constant

The ball's velocity is $\frac{dh}{dt} = 20 - 10t$ , and it was thrown from a height of 1.5 m. Find $h$ .

$h = \int (20 - 10t)\,dt = 20t - 5t^2 + C$ .
At $t = 0$ , $h = 1.5$ , so $C = 1.5$ .

Answer

h = 20t - 5t^2 + 1.5

Watch out

Check any answer by differentiating it. If the derivative is not the integrand, the answer is wrong, and many wrong options fail this test in one line.

Integration by substitution

How does integration by substitution work?

Choose a new variable $t$ for an inner expression whose derivative also appears in the integrand, up to a constant factor. Writing $dt$ for that derivative times $dx$ turns the integral into a standard form in $t$ ; integrate it, then put the inner expression back for $t$ .

Example 3 · Substituting the inside

Evaluate $\int x^2(x^3 + 2)^5\,dx$ .

Put $t = x^3 + 2$ . Then $dt = 3x^2\,dx$ , so $x^2\,dx = \frac{dt}{3}$ .
The integral becomes $\frac{1}{3}\int t^5\,dt = \frac{t^6}{18}$ .

Answer

\frac{(x^3 + 2)^6}{18} + C

Two patterns are worth recognising on sight. When the numerator is the derivative of the denominator, the answer is a logarithm: $\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$ , which gives $\int \tan x\,dx = \log|\sec x| + C$ and $\int \cot x\,dx = \log|\sin x| + C$ . And a linear expression inside only divides by its coefficient: $\int e^{5x - 1}\,dx = \frac{e^{5x - 1}}{5} + C$ .

Watch out

The constant from $dt$ is the usual slip: forgetting the $\frac{1}{3}$ above makes the answer three times too large.

Integration with trigonometric identities

A power or a product of trigonometric functions has no standard form of its own, but an identity turns it into one. The ones used most are:

Identities that make integrals standard

\sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2}, \qquad \tan^2 x = \sec^2 x - 1

2\sin A\cos B = \sin(A + B) + \sin(A - B), \qquad 2\cos A\cos B = \cos(A + B) + \cos(A - B)

\sin^3 x = \frac{3\sin x - \sin 3x}{4}

Example 4 · A product of cosines

Evaluate $\int \cos 4x\cos 2x\,dx$ .

$\cos 4x\cos 2x = \frac{1}{2}(\cos 6x + \cos 2x)$ .
$\frac{1}{2}\left(\frac{\sin 6x}{6} + \frac{\sin 2x}{2}\right)$ .

Answer

\frac{\sin 6x}{12} + \frac{\sin 2x}{4} + C

Integrals of particular functions

What are the standard integrals of 1/(x² + a²) and similar forms?

Six forms come up again and again, and each can be checked by differentiating. With a plus sign and no root, the answer is an inverse tangent; with a minus sign, a logarithm of a ratio; under a root, an inverse sine or a logarithm with $x$ plus the root.

Six particular integrals (add C)

\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}, \qquad \int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right|, \qquad \int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right|

\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a}, \qquad \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right|, \qquad \int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right|

A quadratic $ax^2 + bx + c$ in the denominator is brought to one of these by completing the square.

Example 5 · Completing the square

Evaluate $\int \frac{dx}{x^2 - 6x + 13}$ .

$x^2 - 6x + 13 = (x - 3)^2 + 4$ .
With $t = x - 3$ and $a = 2$ : $\int \frac{dt}{t^2 + 4} = \frac{1}{2}\tan^{-1}\frac{t}{2}$ .

Answer

\frac{1}{2}\tan^{-1}\frac{x - 3}{2} + C

Watch out

Only the $\tan^{-1}$ and the two logarithm-of-a-ratio forms carry a factor in front. The $\sin^{-1}$ form has none.

Integration by partial fractions

How do you integrate using partial fractions?

Split a rational function whose denominator factorises into simpler fractions, one for each factor, find their constants by putting in the roots, and integrate each fraction to a logarithm. If the numerator's degree is not lower than the denominator's, divide first.

Example 6 · Two linear factors

Evaluate $\int \frac{x + 5}{(x + 1)(x + 3)}\,dx$ .

Write $\frac{x + 5}{(x + 1)(x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 3}$ , so $x + 5 = A(x + 3) + B(x + 1)$ .
$x = -1$ gives $4 = 2A$ , so $A = 2$ ; $x = -3$ gives $2 = -2B$ , so $B = -1$ .

Answer

2\log|x + 1| - \log|x + 3| + C

A repeated factor needs one fraction for each power: $\frac{1}{x(x - 2)^2} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}$ , and the last one integrates to a fraction, not a logarithm.

Integration by parts

When do you use integration by parts?

Use it for a product of two different kinds of function, such as $x$ times $e^x$ , $\sin x$ or $\log x$ . Take one factor as $u$ , to be differentiated, and the rest as $dv$ , to be integrated; then $\int u\,dv = uv - \int v\,du$ , where the new integral should be easier than the old one.

To choose $u$ , take whichever factor comes first in the order ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. So $u = \log x$ in $\int x^2\log x\,dx$ , and $u = x$ in $\int xe^x\,dx$ .

Example 7 · By parts

Evaluate $\int x\sin x\,dx$ .

$u = x$ and $dv = \sin x\,dx$ , so $du = dx$ and $v = -\cos x$ .
$uv - \int v\,du = -x\cos x + \int \cos x\,dx$ .

Answer

-x\cos x + \sin x + C

Watch out

The second integral is subtracted. Many wrong options add it, and differentiating them shows the slip at once.

A single function with no obvious second factor, like $\log x$ or $\tan^{-1} x$ , is integrated by parts with $dv = 1 \cdot dx$ : $\int \log x\,dx = x\log x - x + C$ .

Example 8 · A function and its derivative beside the exponential

Evaluate $\int e^x(\sin x + \cos x)\,dx$ .

$\cos x$ is the derivative of $\sin x$ , so the integrand is $e^x\left(f(x) + f'(x)\right)$ with $f(x) = \sin x$ .
$\frac{d}{dx}\left(e^x\sin x\right) = e^x\sin x + e^x\cos x$ , exactly the integrand.

Answer

e^x\sin x + C

How CUET asks this chapter

Standard forms: term-by-term integrals, often with trigonometric or logarithmic terms, or an antiderivative fixed by a given value.
Substitution: an integrand with an inner expression and its derivative, including the $\frac{f'}{f}$ form.
Identities: $\sin^2 x$ , $\tan^2 x$ , $\sin^3 x$ or a product such as $\sin 3x\cos x$ .
Particular forms: $\frac{1}{x^2 \pm a^2}$ and the root forms, sometimes after completing the square, and often as match-the-list.
Partial fractions and by parts: a split into fractions, one constant of a split, or a product such as $xe^x$ , with $e^x(f + f')$ a favourite.

Exam shortcuts

Differentiate the options: the right one gives back the integrand.
$\int f(ax + b)\,dx = \frac{1}{a}F(ax + b) + C$ .
$\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$ .
$\int e^x\left(f(x) + f'(x)\right)dx = e^x f(x) + C$ .
$\int \log x\,dx = x\log x - x + C$ ; $\int \tan x\,dx = \log|\sec x| + C$ .
ILATE picks $u$ for integration by parts.

Common mistakes

Differentiating instead of integrating.
Forgetting the constant from $dt$ , or the $\frac{1}{a}$ from a linear inside.
Writing $\log x$ for $\int \frac{dx}{x}$ instead of $\log|x|$ .
Mixing up the particular forms: $\tan^{-1}$ belongs to $x^2 + a^2$ , $\sin^{-1}$ to $\sqrt{a^2 - x^2}$ .
Adding the second integral in integration by parts instead of subtracting it.
Integrating an improper fraction without dividing first.

Quick revision

Basics

$\int f\,dx = F + C$ when $F' = f$
Sums and constant factors split; products do not

Methods

Substitution: $t$ = an inside whose derivative is present
Identities for powers and products of sine and cosine
Partial fractions for a factorised denominator

By parts

$\int u\,dv = uv - \int v\,du$ ; choose $u$ by ILATE
$\int e^x(f + f')\,dx = e^x f + C$

Put limits on an integral and it becomes a number, and an area: definite integrals and area.

Next · 10Quick revision

Check your understanding

90 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Integration as the reverse of differentiationQ1 of 90Q1

Integration as the reverse of differentiation

15 questions

Question 1

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Evaluate $\int \left(3 x^{2} + 4 x - 5\right)\,dx$ .

6 x + 4 + C

Correct answerYour answer

x^{3} + 2 x^{2} - 5 x + C

Correct answerYour answer

3 x^{3} + 4 x^{2} - 5 x + C

Correct answerYour answer

- x^{3} - 2 x^{2} + 5 x + C

Correct answerYour answer

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AnswerB · $x^{3} + 2 x^{2} - 5 x + C$

Integrate term by term with the standard forms: $\int x^n\,dx = \frac{x^{n+1}}{n+1}$ , $\int \frac{1}{x}\,dx = \log|x|$ , $\int e^x\,dx = e^x$ .
$\int \left(3 x^{2} + 4 x - 5\right)\,dx = x^{3} + 2 x^{2} - 5 x + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

$6 x + 4$ differentiates instead of integrating.

Question 2

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If $F'(x) = 3x^2$ and $F(1) = 4$ , what is $F(x)$ ?

x^{3} + 4

Correct answerYour answer

x^{3} + 3

Correct answerYour answer

x^{3}

Correct answerYour answer

6 x - 2

Correct answerYour answer

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AnswerB · $x^{3} + 3$

$F(x) = \int 3x^2\,dx = x^3 + C$ .
$F(1) = 1 + C = 4$ gives $C = 3$ , so $F(x) = x^3 + 3$ .

Worth noting

$x^3 + 4$ gives $F(1) = 5$ : the 4 is the value at 1, not the constant.

Question 3

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Evaluate $\int \left(3 e^{x} + \frac{4}{x}\right)\,dx$ .

3 e^{x} + 4 \log\left|x\right| + C

Correct answerYour answer

3 e^{x} - \frac{4}{x^{2}} + C

Correct answerYour answer

3 x e^{x} + 4 + C

Correct answerYour answer

- 3 e^{x} - 4 \log\left|x\right| + C

Correct answerYour answer

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AnswerA · $3 e^{x} + 4 \log\left|x\right| + C$

Integrate term by term with the standard forms: $\int x^n\,dx = \frac{x^{n+1}}{n+1}$ , $\int \frac{1}{x}\,dx = \log|x|$ , $\int e^x\,dx = e^x$ .
$\int \left(3 e^{x} + \frac{4}{x}\right)\,dx = 3 e^{x} + 4 \log\left|x\right| + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

$3 e^{x} - \frac{4}{x^{2}}$ differentiates instead of integrating.

Question 4

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Evaluate $\int 2^{x}\,dx$ .

2^{x}\log 2 + C

Correct answerYour answer

\frac{2^{x + 1}}{x + 1} + C

Correct answerYour answer

\frac{2^{x}}{\log 2} + C

Correct answerYour answer

x \cdot 2^{x - 1} + C

Correct answerYour answer

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AnswerC · $\frac{2^{x}}{\log 2} + C$

$\frac{d}{dx}a^{x} = a^{x}\log a$ , so $\int a^{x}\,dx = \frac{a^{x}}{\log a} + C$ .
With $a = 2$ : $\frac{2^{x}}{\log 2} + C$ .

Worth noting

$2^{x}\log 2$ is the derivative of $2^{x}$ , not its integral.

Question 5

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If $F'(x) = 4x^3 - 6x$ and $F(2) = 3$ , what is $F(x)$ ?

x^{4} - 3 x^{2} + 3

Correct answerYour answer

x^{4} - 3 x^{2} + 1

Correct answerYour answer

12 x^{2} - 6

Correct answerYour answer

x^{4} - 3 x^{2} - 1

Correct answerYour answer

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AnswerD · $x^{4} - 3 x^{2} - 1$

$F(x) = \int (4x^3 - 6x)\,dx = x^4 - 3x^2 + C$ .
$F(2) = 16 - 12 + C = 3$ , so $C = -1$ .

Worth noting

$+3$ copies $F(2)$ into the constant without working out $x^4 - 3x^2$ at $x = 2$ .

Question 6

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Evaluate $\int \left(\operatorname{cosec}^{2} x + \sec^{2} x\right)\,dx$ .

\tan x + \cot x + C

Correct answerYour answer

\sec x\tan x + \operatorname{cosec} x\cot x + C

Correct answerYour answer

\tan x - \cot x + C

Correct answerYour answer

\cot x - \tan x + C

Correct answerYour answer

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AnswerC · $\tan x - \cot x + C$

$\frac{d}{dx}\tan x = \sec^2 x$ and $\frac{d}{dx}\cot x = -\operatorname{cosec}^2 x$ .
So $\int \sec^2 x\,dx = \tan x$ and $\int \operatorname{cosec}^2 x\,dx = -\cot x$ : the sum is $\tan x - \cot x + C$ .

Worth noting

$\tan x + \cot x$ misses the minus sign: the derivative of $\cot x$ is negative.

Question 7

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Which of these is an antiderivative of $2x\cos(x^2)$ ?

- \sin{\left(x^{2} \right)}

Correct answerYour answer

x^{2} \sin{\left(x^{2} \right)}

Correct answerYour answer

\cos{\left(x^{2} \right)}

Correct answerYour answer

\sin{\left(x^{2} \right)}

Correct answerYour answer

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AnswerD · $\sin{\left(x^{2} \right)}$

An antiderivative $F$ is any function with $F'(x)$ equal to the given function.
$\frac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x$ , which is the given function.

Worth noting

$-\sin(x^2)$ differentiates to $-2x\cos(x^2)$ : the sign is wrong.

Question 8

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Assertion (A): $\int \frac{dx}{x} = \log|x| + C$ on every interval that does not contain 0, negative $x$ included.

Reason (R): $\frac{d}{dx}\log|x| = \frac{1}{x}$ for every $x \ne 0$ , negative $x$ included.

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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AnswerA · Both A and R are true, and R is the correct explanation of A.

For $x < 0$ , $\log|x| = \log(-x)$ , whose derivative is $\frac{-1}{-x} = \frac{1}{x}$ .
So R is true, and it is exactly why A holds for negative $x$ as well: A is true and R explains it.

Question 9

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Which of the following statements are true?

I. $\int (f + g)\,dx = \int f\,dx + \int g\,dx$ .

II. $\int kf\,dx = k\int f\,dx$ for a constant $k$ .

III. $\int fg\,dx = \int f\,dx \cdot \int g\,dx$ .

IV. Two antiderivatives of the same function differ by a constant.

All four: I, II, III and IVCorrect answerYour answerI and II onlyCorrect answerYour answerI, II and IV onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answer

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AnswerC · I, II and IV only

I and II are the two properties of 7.2.1, and IV holds because two functions with the same derivative differ by a constant.
III is false: with $f = g = x$ , $\int x^2\,dx = \frac{x^3}{3}$ , but $\frac{x^2}{2} \cdot \frac{x^2}{2} = \frac{x^4}{4}$ .

Question 10

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Evaluate $\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)\,dx$ .

\frac{1}{2 \sqrt{x}} - \frac{1}{2 x^{\frac{3}{2}}} + C

Correct answerYour answer

x^{\frac{3}{2}} + \sqrt{x} + C

Correct answerYour answer

- \frac{2 x^{\frac{3}{2}}}{3} - 2 \sqrt{x} + C

Correct answerYour answer

\frac{2 x^{\frac{3}{2}}}{3} + 2 \sqrt{x} + C

Correct answerYour answer

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AnswerD · $\frac{2 x^{\frac{3}{2}}}{3} + 2 \sqrt{x} + C$

Integrate term by term with the standard forms: $\int x^n\,dx = \frac{x^{n+1}}{n+1}$ , $\int \frac{1}{x}\,dx = \log|x|$ , $\int e^x\,dx = e^x$ .
$\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)\,dx = \frac{2 x^{\frac{3}{2}}}{3} + 2 \sqrt{x} + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

$\frac{1}{2 \sqrt{x}} - \frac{1}{2 x^{\frac{3}{2}}}$ differentiates instead of integrating.

Question 11

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Evaluate $\int \left(x + \frac{1}{x}\right)^{2}\,dx$ .

\frac{1}{3}\left(x + \frac{1}{x}\right)^{3} + C

Correct answerYour answer

\frac{x^{3}}{3} + 2x + \frac{1}{x} + C

Correct answerYour answer

\frac{x^{3}}{3} - \frac{1}{x} + C

Correct answerYour answer

\frac{x^{3}}{3} + 2x - \frac{1}{x} + C

Correct answerYour answer

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AnswerD · $\frac{x^{3}}{3} + 2x - \frac{1}{x} + C$

Expand first: $\left(x + \frac{1}{x}\right)^{2} = x^{2} + 2 + \frac{1}{x^{2}}$ .
$\int \left(x^{2} + 2 + x^{-2}\right)dx = \frac{x^{3}}{3} + 2x - \frac{1}{x} + C$ .

Worth noting

$\frac{1}{3}\left(x + \frac{1}{x}\right)^{3}$ treats the bracket like $x$ ; its derivative carries an extra factor $1 - \frac{1}{x^{2}}$ .

Question 12

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Evaluate $\int \frac{x^{3} - 1}{x - 1}\,dx$ .

\frac{x^{3}}{3} + \frac{x^{2}}{2} + x + C

Correct answerYour answer

x^{3} + x^{2} + x + C

Correct answerYour answer

\frac{x^{3}}{3} + \frac{x^{2}}{2} + C

Correct answerYour answer

\frac{x^{3}}{3} - \frac{x^{2}}{2} + x + C

Correct answerYour answer

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AnswerA · $\frac{x^{3}}{3} + \frac{x^{2}}{2} + x + C$

Divide first: $x^{3} - 1 = (x - 1)(x^{2} + x + 1)$ , so the integrand is $x^{2} + x + 1$ .
$\int (x^{2} + x + 1)\,dx = \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + C$ .

Worth noting

$x^{3} + x^{2} + x$ raises each power but forgets to divide by the new power.

Question 13

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A curve passes through $(1, 6)$ , and its slope at every point $(x, y)$ is $2x + 3$ . What is its equation?

y = x^{2} + 3x + 6

Correct answerYour answer

y = x^{2} + 3x

Correct answerYour answer

y = x^{2} + 3x + 2

Correct answerYour answer

y = 2x^{2} + 3x + 1

Correct answerYour answer

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AnswerC · $y = x^{2} + 3x + 2$

$y = \int (2x + 3)\,dx = x^{2} + 3x + C$ .
Through $(1, 6)$ : $6 = 1 + 3 + C$ , so $C = 2$ .

Worth noting

$+6$ takes the point's $y$ as the constant; the constant is what is left after $x^{2} + 3x$ at $x = 1$ .

Question 14

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Evaluate $\int \frac{\sin x}{\cos^{2} x}\,dx$ .

-\sec x + C

Correct answerYour answer

\sec x + C

Correct answerYour answer

\tan x + C

Correct answerYour answer

\sec x\tan x + C

Correct answerYour answer

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AnswerB · $\sec x + C$

$\frac{\sin x}{\cos^{2} x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x\tan x$ .
$\int \sec x\tan x\,dx = \sec x + C$ .

Worth noting

$-\sec x$ gets the sign wrong: $\frac{d}{dx}\sec x = +\sec x\tan x$ .

Question 15

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Which of the following statements are true?

I. $\int e^{2x}\,dx = \frac{e^{2x}}{2} + C$ .

II. $\int \frac{1}{x^{2}}\,dx = \log x^{2} + C$ .

III. $\int k f(x)\,dx = k\int f(x)\,dx$ for a constant $k$ .

IV. $\int f(x)g(x)\,dx = \int f(x)\,dx \cdot \int g(x)\,dx$ .

I and III onlyCorrect answerYour answerI, III and IV onlyCorrect answerYour answerI onlyCorrect answerYour answerI, II and III onlyCorrect answerYour answer

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AnswerA · I and III only

I: $\frac{d}{dx}\frac{e^{2x}}{2} = e^{2x}$ . III: a constant factor comes out of an integral.
II is false: $\int x^{-2}\,dx = -\frac{1}{x} + C$ . IV is false: there is no product rule for integrals (try $f = g = x$ ).

Worth noting

$\log$ appears only for the power $-1$ ; every other power, $-2$ included, follows $\frac{x^{n + 1}}{n + 1}$ .

Integration by substitution

18 questions

Question 16

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Evaluate $\int 2 x \left(x^{2} + 1\right)^{4}\,dx$ .

\frac{2(x^2 + 1)^5}{5} + C

Correct answerYour answer

\frac{(x^2 + 1)^5}{5} + C

Correct answerYour answer

-\frac{(x^2 + 1)^5}{5} + C

Correct answerYour answer

2 x^{2} \left(x^{2} + 1\right)^{4} + C

Correct answerYour answer

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AnswerB · $\frac{(x^2 + 1)^5}{5} + C$

Put $t = x^2 + 1$ ; then $dt$ is the rest of the integrand, up to a constant.
The integral becomes a standard form in $t$ ; putting $t = x^2 + 1$ back gives $\frac{(x^2 + 1)^5}{5} + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

Getting the constant from $dt$ wrong multiplies or divides the answer: check by differentiating.

Question 17

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Evaluate $\int \cos{\left(3 x + 2 \right)}\,dx$ .

\frac{\sin{\left(3 x + 2 \right)}}{3} + C

Correct answerYour answer

\sin{\left(3 x + 2 \right)} + C

Correct answerYour answer

- \frac{\sin{\left(3 x + 2 \right)}}{3} + C

Correct answerYour answer

x \cos{\left(3 x + 2 \right)} + C

Correct answerYour answer

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AnswerA · $\frac{\sin{\left(3 x + 2 \right)}}{3} + C$

Put $t = 3x + 2$ ; then $dt$ is the rest of the integrand, up to a constant.
The integral becomes a standard form in $t$ ; putting $t = 3x + 2$ back gives $\frac{\sin{\left(3 x + 2 \right)}}{3} + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

Getting the constant from $dt$ wrong multiplies or divides the answer: check by differentiating.

Question 18

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Which substitution turns $\int x^2e^{x^3}\,dx$ into a standard form?

t = x^2

Correct answerYour answer

t = e^x

Correct answerYour answer

t = x^3

Correct answerYour answer

t = x^2e^{x^3}

Correct answerYour answer

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AnswerC · $t = x^3$

With $t = x^3$ , $dt = 3x^2\,dx$ , so $x^2\,dx = \frac{dt}{3}$ , which is the rest of the integrand.
The integral becomes $\frac{1}{3}\int e^t\,dt = \frac{1}{3}e^{x^3} + C$ .

Worth noting

Choose $t$ so that its derivative, up to a constant, is the factor left over.

Question 19

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Evaluate $\int \cot x\,dx$ .

-\log\left|\sin x\right| + C

Correct answerYour answer

\log\left|\cos x\right| + C

Correct answerYour answer

\log\left|\sin x\right| + C

Correct answerYour answer

-\operatorname{cosec}^{2} x + C

Correct answerYour answer

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AnswerC · $\log\left|\sin x\right| + C$

$\cot x = \frac{\cos x}{\sin x}$ : the top is the derivative of the bottom.
So $\int \cot x\,dx = \log|\sin x| + C$ .

Worth noting

$-\log|\sin x|$ mixes it up with $\int \tan x\,dx = -\log|\cos x|$ , which does carry a minus.

Question 20

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Evaluate $\int e^{3x + 1}\,dx$ .

e^{3x + 1} + C

Correct answerYour answer

3e^{3x + 1} + C

Correct answerYour answer

\frac{e^{3x + 1}}{3x + 1} + C

Correct answerYour answer

\frac{1}{3}e^{3x + 1} + C

Correct answerYour answer

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AnswerD · $\frac{1}{3}e^{3x + 1} + C$

Put $t = 3x + 1$ , $dt = 3\,dx$ : $\frac{1}{3}\int e^{t}\,dt$ .
$= \frac{1}{3}e^{3x + 1} + C$ .

Worth noting

$e^{3x + 1}$ alone differentiates to $3e^{3x + 1}$ , three times too much.

Question 21

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Evaluate $\int x e^{x^{2}}\,dx$ .

e^{x^{2}} + C

Correct answerYour answer

- \frac{e^{x^{2}}}{2} + C

Correct answerYour answer

x^{2} e^{x^{2}} + C

Correct answerYour answer

\frac{e^{x^{2}}}{2} + C

Correct answerYour answer

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AnswerD · $\frac{e^{x^{2}}}{2} + C$

Put $t = x^2$ ; then $dt$ is the rest of the integrand, up to a constant.
The integral becomes a standard form in $t$ ; putting $t = x^2$ back gives $\frac{e^{x^{2}}}{2} + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

Getting the constant from $dt$ wrong multiplies or divides the answer: check by differentiating.

Question 22

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Evaluate $\int \tan x\,dx$ .

2\log\left|\sec x\right| + C

Correct answerYour answer

-\log\left|\sec x\right| + C

Correct answerYour answer

\log\left|\sec x\right| + C

Correct answerYour answer

x \tan x + C

Correct answerYour answer

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AnswerC · $\log\left|\sec x\right| + C$

Put $t = \cos x$ ; then $dt$ is the rest of the integrand, up to a constant.
The integral becomes a standard form in $t$ ; putting $t = \cos x$ back gives $\log\left|\sec x\right| + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

Getting the constant from $dt$ wrong multiplies or divides the answer: check by differentiating.

Question 23

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Evaluate $\int \frac{2 x + 3}{x^{2} + 3 x + 5}\,dx$ .

2 \log\left|x^{2} + 3 x + 5\right| + C

Correct answerYour answer

\log\left|x^{2} + 3 x + 5\right| + C

Correct answerYour answer

- \log\left|x^{2} + 3 x + 5\right| + C

Correct answerYour answer

\frac{x \left(2 x + 3\right)}{x^{2} + 3 x + 5} + C

Correct answerYour answer

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AnswerB · $\log\left|x^{2} + 3 x + 5\right| + C$

Put $t = x^2 + 3x + 5$ ; then $dt$ is the rest of the integrand, up to a constant.
The integral becomes a standard form in $t$ ; putting $t = x^2 + 3x + 5$ back gives $\log\left|x^{2} + 3 x + 5\right| + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

Getting the constant from $dt$ wrong multiplies or divides the answer: check by differentiating.

Question 24

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Evaluate $\int e^{\sin x} \cos x\,dx$ .

2 e^{\sin x} + C

Correct answerYour answer

- e^{\sin x} + C

Correct answerYour answer

x e^{\sin x} \cos x + C

Correct answerYour answer

e^{\sin x} + C

Correct answerYour answer

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AnswerD · $e^{\sin x} + C$

Put $t = \sin x$ ; then $dt$ is the rest of the integrand, up to a constant.
The integral becomes a standard form in $t$ ; putting $t = \sin x$ back gives $e^{\sin x} + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

Getting the constant from $dt$ wrong multiplies or divides the answer: check by differentiating.

Question 25

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Evaluate $\int \frac{\sec^{2} x}{\tan x + 1}\,dx$ .

\frac{(1 + \tan x)^2}{2} + C

Correct answerYour answer

\log\left|1 + \tan x\right| + C

Correct answerYour answer

\tan x + \log\left|\tan x\right| + C

Correct answerYour answer

-\frac{1}{1 + \tan x} + C

Correct answerYour answer

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AnswerB · $\log\left|1 + \tan x\right| + C$

The numerator $\sec^2 x$ is the derivative of the denominator $1 + \tan x$ .
With $t = 1 + \tan x$ : $\int \frac{dt}{t} = \log|t| = \log|1 + \tan x| + C$ .

Worth noting

$-\frac{1}{1 + \tan x}$ would be the answer for $\frac{\sec^2 x}{(1 + \tan x)^2}$ , with the denominator squared.

Question 26

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Evaluate $\int \frac{\log x}{x}\,dx$ .

(\log x)^{2} + C

Correct answerYour answer

\log(\log x) + C

Correct answerYour answer

\frac{\log x}{x^{2}} + C

Correct answerYour answer

\frac{(\log x)^{2}}{2} + C

Correct answerYour answer

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AnswerD · $\frac{(\log x)^{2}}{2} + C$

Put $t = \log x$ , so $dt = \frac{dx}{x}$ .
$\int t\,dt = \frac{t^{2}}{2} = \frac{(\log x)^{2}}{2} + C$ .

Worth noting

$(\log x)^{2}$ forgets to divide by 2; its derivative is $\frac{2\log x}{x}$ .

Question 27

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Evaluate $\int \sec x\,dx$ .

\sec x\tan x + C

Correct answerYour answer

\log\left|\sec x + \tan x\right| + C

Correct answerYour answer

\log\left|\sec x - \tan x\right| + C

Correct answerYour answer

\log\left|\cos x\right| + C

Correct answerYour answer

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AnswerB · $\log\left|\sec x + \tan x\right| + C$

Multiply top and bottom by $\sec x + \tan x$ : $\int \frac{\sec^{2} x + \sec x\tan x}{\sec x + \tan x}\,dx$ .
The top is now the derivative of the bottom, so the integral is $\log|\sec x + \tan x| + C$ .

Worth noting

$\sec x\tan x$ is the derivative of $\sec x$ , not its integral.

Question 28

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Evaluate $\int x\sqrt{x^{2} + 4}\,dx$ .

\frac{2}{3}\left(x^{2} + 4\right)^{\frac{3}{2}} + C

Correct answerYour answer

\frac{1}{3}\left(x^{2} + 4\right)^{\frac{3}{2}} + C

Correct answerYour answer

\frac{1}{6}\left(x^{2} + 4\right)^{\frac{3}{2}} + C

Correct answerYour answer

\frac{x^{2}}{3}\left(x^{2} + 4\right)^{\frac{3}{2}} + C

Correct answerYour answer

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AnswerB · $\frac{1}{3}\left(x^{2} + 4\right)^{\frac{3}{2}} + C$

Put $t = x^{2} + 4$ , so $dt = 2x\,dx$ and $x\,dx = \frac{dt}{2}$ .
$\frac{1}{2}\int t^{1/2}\,dt = \frac{1}{2} \cdot \frac{2}{3}t^{3/2} = \frac{1}{3}(x^{2} + 4)^{3/2} + C$ .

Worth noting

$\frac{2}{3}$ forgets the $\frac{1}{2}$ from $x\,dx = \frac{dt}{2}$ .

Question 29

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Evaluate $\int \sin x\cos x\,dx$ .

\frac{\sin^{2} x}{2} + C

Correct answerYour answer

\sin^{2} x + C

Correct answerYour answer

\frac{\cos^{2} x}{2} + C

Correct answerYour answer

-\frac{\cos 2x}{2} + C

Correct answerYour answer

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AnswerA · $\frac{\sin^{2} x}{2} + C$

Put $t = \sin x$ , $dt = \cos x\,dx$ : $\int t\,dt = \frac{t^{2}}{2}$ .
$= \frac{\sin^{2} x}{2} + C$ . (Through $\sin x\cos x = \frac{1}{2}\sin 2x$ one gets $-\frac{\cos 2x}{4} + C$ , the same up to the constant.)

Worth noting

$\frac{\cos^{2} x}{2}$ has the sign wrong: its derivative is $-\sin x\cos x$ .

Question 30

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Evaluate $\int \tan^{3} x\sec^{2} x\,dx$ .

\frac{\sec^{3} x}{3} + C

Correct answerYour answer

\frac{\tan^{4} x}{4} + C

Correct answerYour answer

\tan^{4} x + C

Correct answerYour answer

\frac{\tan^{4} x\sec^{3} x}{12} + C

Correct answerYour answer

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AnswerB · $\frac{\tan^{4} x}{4} + C$

Put $t = \tan x$ , $dt = \sec^{2} x\,dx$ .
$\int t^{3}\,dt = \frac{t^{4}}{4} = \frac{\tan^{4} x}{4} + C$ .

Worth noting

$\frac{\tan^{4} x\sec^{3} x}{12}$ integrates the two factors separately and multiplies, which is not how integrals work.

Question 31

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Evaluate $\int \frac{\cos{\left(\sqrt{x} \right)}}{\sqrt{x}}\,dx$ .

2 \sin{\left(\sqrt{x} \right)} + C

Correct answerYour answer

4 \sin{\left(\sqrt{x} \right)} + C

Correct answerYour answer

- 2 \sin{\left(\sqrt{x} \right)} + C

Correct answerYour answer

\sqrt{x} \cos{\left(\sqrt{x} \right)} + C

Correct answerYour answer

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AnswerA · $2 \sin{\left(\sqrt{x} \right)} + C$

Put $t = \sqrt{x}$ ; then $dt$ is the rest of the integrand, up to a constant.
The integral becomes a standard form in $t$ ; putting $t = \sqrt{x}$ back gives $2 \sin{\left(\sqrt{x} \right)} + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

Getting the constant from $dt$ wrong multiplies or divides the answer: check by differentiating.

Question 32

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Evaluate $\int \frac{dx}{x\log x}$ .

\log\left|\log x\right| + C

Correct answerYour answer

\frac{(\log x)^{2}}{2} + C

Correct answerYour answer

\frac{1}{\log x} + C

Correct answerYour answer

\log x \cdot \log\left|x\right| + C

Correct answerYour answer

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AnswerA · $\log\left|\log x\right| + C$

Put $t = \log x$ , so $dt = \frac{dx}{x}$ and the integral becomes $\int \frac{dt}{t}$ .
$= \log|t| + C = \log|\log x| + C$ .

Worth noting

$\frac{(\log x)^{2}}{2}$ is $\int \frac{\log x}{x}\,dx$ : here $\log x$ is in the denominator.

Question 33

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Evaluate $\int \frac{dx}{1 + e^{-x}}$ .

\log\left(1 + e^{-x}\right) + C

Correct answerYour answer

x - \log\left(1 + e^{x}\right) + C

Correct answerYour answer

\log\left(1 + e^{x}\right) + C

Correct answerYour answer

x + e^{-x} + C

Correct answerYour answer

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AnswerC · $\log\left(1 + e^{x}\right) + C$

Multiply top and bottom by $e^{x}$ : $\frac{1}{1 + e^{-x}} = \frac{e^{x}}{e^{x} + 1}$ .
The top is the derivative of the bottom: $\log(1 + e^{x}) + C$ .

Worth noting

$\log(1 + e^{-x})$ takes the bottom as it stands, but its derivative is $-e^{-x}$ , not the top.

Integration with trigonometric identities

12 questions

Question 34

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Evaluate $\int \tan^{2} x\,dx$ .

\frac{\tan^3 x}{3} + C

Correct answerYour answer

-\left(\tan x - x\right) + C

Correct answerYour answer

x \tan^{2} x + C

Correct answerYour answer

\tan x - x + C

Correct answerYour answer

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AnswerD · $\tan x - x + C$

Rewrite first: $\tan^2 x = \sec^2 x - 1$ .
Then each term is standard: $\tan x - x + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

A power of a trig function has no direct standard form: the identity is what makes it integrable.

Question 35

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Evaluate $\int \frac{\sin 2x}{\sin x}\,dx$ .

-2\cos x + C

Correct answerYour answer

2\cos x + C

Correct answerYour answer

2\sin x + C

Correct answerYour answer

\sin^{2} x + C

Correct answerYour answer

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AnswerC · $2\sin x + C$

$\sin 2x = 2\sin x\cos x$ , so the integrand is $2\cos x$ .
$\int 2\cos x\,dx = 2\sin x + C$ .

Worth noting

$-2\cos x$ integrates $2\sin x$ instead of $2\cos x$ .

Question 36

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Evaluate $\int \sin^{2} x\,dx$ .

\frac{x}{2} - \frac{\sin 2x}{4} + C

Correct answerYour answer

\frac{x}{2} + \frac{\sin 2x}{4} + C

Correct answerYour answer

-\left(\frac{x}{2} - \frac{\sin 2x}{4}\right) + C

Correct answerYour answer

x \sin^{2} x + C

Correct answerYour answer

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AnswerA · $\frac{x}{2} - \frac{\sin 2x}{4} + C$

Rewrite first: $\sin^2 x = \frac{1 - \cos 2x}{2}$ .
Then each term is standard: $\frac{x}{2} - \frac{\sin 2x}{4} + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

A power of a trig function has no direct standard form: the identity is what makes it integrable.

Question 37

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Evaluate $\int \frac{1}{\cos 2x + 1}\,dx$ .

2\tan x + C

Correct answerYour answer

-\left(\frac{\tan x}{2}\right) + C

Correct answerYour answer

\frac{\tan x}{2} + C

Correct answerYour answer

\frac{x}{\cos 2x + 1} + C

Correct answerYour answer

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AnswerC · $\frac{\tan x}{2} + C$

Rewrite first: $1 + \cos 2x = 2\cos^2 x$ .
Then each term is standard: $\frac{\tan x}{2} + C$ .
Check: differentiating the answer gives back the integrand.

Worth noting

A power of a trig function has no direct standard form: the identity is what makes it integrable.

Question 38

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Evaluate $\int \cos^{2} 2x\,dx$ .

\frac{x}{2} + \frac{\sin 4x}{4} + C

Correct answerYour answer

\frac{x}{2} - \frac{\sin 4x}{8} + C

Correct answerYour answer

\frac{\cos^{3} 2x}{6} + C

Correct answerYour answer

\frac{x}{2} + \frac{\sin 4x}{8} + C

Correct answerYour answer

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AnswerD · $\frac{x}{2} + \frac{\sin 4x}{8} + C$

$\cos^{2} 2x = \frac{1 + \cos 4x}{2}$ .
$\int \frac{1 + \cos 4x}{2}\,dx = \frac{x}{2} + \frac{\sin 4x}{8} + C$ .

Worth noting

$\frac{\sin 4x}{4}$ drops the $\frac{1}{2}$ in front of $\cos 4x$ .

Question 39

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Evaluate $\int (\sin x + \cos x)^{2}\,dx$ .

\frac{(\sin x + \cos x)^{3}}{3} + C

Correct answerYour answer

x - \frac{\cos 2x}{2} + C

Correct answerYour answer

x + \frac{\cos 2x}{2} + C

Correct answerYour answer

x - \cos 2x + C

Correct answerYour answer

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AnswerB · $x - \frac{\cos 2x}{2} + C$

$(\sin x + \cos x)^{2} = \sin^{2} x + \cos^{2} x + 2\sin x\cos x = 1 + \sin 2x$ .
$\int (1 + \sin 2x)\,dx = x - \frac{\cos 2x}{2} + C$ .

Worth noting

Raising the bracket to the power 3 treats it like $x$ ; its derivative brings an extra factor $\cos x - \sin x$ .

Question 40

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Evaluate $\int \sin 3x \cos x\,dx$ .

-\frac{\cos 3x}{3}\sin x + C

Correct answerYour answer

-\frac{\cos 4x}{8} - \frac{\cos 2x}{4} + C

Correct answerYour answer

\frac{\cos 4x}{8} + \frac{\cos 2x}{4} + C

Correct answerYour answer

-\frac{\cos 4x}{4} - \frac{\cos 2x}{2} + C

Correct answerYour answer

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AnswerB · $-\frac{\cos 4x}{8} - \frac{\cos 2x}{4} + C$

Product to sum: $\sin 3x\cos x = \frac{1}{2}(\sin 4x + \sin 2x)$ .
$\frac{1}{2}\left(-\frac{\cos 4x}{4} - \frac{\cos 2x}{2}\right) = -\frac{\cos 4x}{8} - \frac{\cos 2x}{4}$ , plus $C$ .

Worth noting

The answer without the $\frac{1}{2}$ of the product-to-sum formula is twice too large.

Question 41

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Evaluate $\int \sin^{3} x\,dx$ .

\frac{\sin^4 x}{4} + C

Correct answerYour answer

-\frac{3\cos x}{4} - \frac{\cos 3x}{12} + C

Correct answerYour answer

-\frac{\cos^4 x}{4} + C

Correct answerYour answer

-\frac{3\cos x}{4} + \frac{\cos 3x}{12} + C

Correct answerYour answer

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AnswerD · $-\frac{3\cos x}{4} + \frac{\cos 3x}{12} + C$

$\sin 3x = 3\sin x - 4\sin^3 x$ , so $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$ .
$\int = \frac{1}{4}\left(-3\cos x + \frac{\cos 3x}{3}\right) = -\frac{3\cos x}{4} + \frac{\cos 3x}{12} + C$ .

Worth noting

$\frac{\sin^4 x}{4}$ treats $\sin x$ like $x$ in the power rule; its derivative has an extra $\cos x$ .

Question 42

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Evaluate $\int \frac{\cos 2x}{(\cos x + \sin x)^2}\,dx$ .

\log\left|\cos x + \sin x\right| + C

Correct answerYour answer

\log\left|\cos x - \sin x\right| + C

Correct answerYour answer

-\frac{1}{\cos x + \sin x} + C

Correct answerYour answer

\frac{\sin 2x}{2} + C

Correct answerYour answer

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AnswerA · $\log\left|\cos x + \sin x\right| + C$

$\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$ , so the integrand is $\frac{\cos x - \sin x}{\cos x + \sin x}$ .
The numerator is the derivative of the denominator, so the integral is $\log|\cos x + \sin x| + C$ .

Worth noting

$\log|\cos x - \sin x|$ puts the wrong factor in the logarithm: it is the denominator that remains.

Question 43

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Evaluate $\int 2\sin 5x\sin x\,dx$ .

\frac{\sin 4x}{4} - \frac{\sin 6x}{6} + C

Correct answerYour answer

\frac{\sin 4x}{4} + \frac{\sin 6x}{6} + C

Correct answerYour answer

\frac{\sin 6x}{6} - \frac{\sin 4x}{4} + C

Correct answerYour answer

\sin 4x - \sin 6x + C

Correct answerYour answer

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AnswerA · $\frac{\sin 4x}{4} - \frac{\sin 6x}{6} + C$

$2\sin A\sin B = \cos(A - B) - \cos(A + B)$ : $2\sin 5x\sin x = \cos 4x - \cos 6x$ .
$\int (\cos 4x - \cos 6x)\,dx = \frac{\sin 4x}{4} - \frac{\sin 6x}{6} + C$ .

Worth noting

A plus sign between the terms uses the cosine product formula, $2\cos A\cos B$ , instead.

Question 44

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Evaluate $\int \cos^{3} x\,dx$ .

\frac{\cos^{4} x}{4} + C

Correct answerYour answer

\sin x + \frac{\sin^{3} x}{3} + C

Correct answerYour answer

\frac{\sin^{3} x}{3} + C

Correct answerYour answer

\sin x - \frac{\sin^{3} x}{3} + C

Correct answerYour answer

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AnswerD · $\sin x - \frac{\sin^{3} x}{3} + C$

$\cos^{3} x = (1 - \sin^{2} x)\cos x$ . Put $t = \sin x$ , $dt = \cos x\,dx$ .
$\int (1 - t^{2})\,dt = t - \frac{t^{3}}{3} = \sin x - \frac{\sin^{3} x}{3} + C$ .

Worth noting

$\frac{\cos^{4} x}{4}$ uses the power rule on $\cos x$ , but $dx$ is not $d(\cos x)$ .

Question 45

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Evaluate $\int \sin^{2} x\cos^{2} x\,dx$ .

\frac{x}{8} - \frac{\sin 4x}{32} + C

Correct answerYour answer

\frac{x}{8} + \frac{\sin 4x}{32} + C

Correct answerYour answer

\frac{x}{4} - \frac{\sin 4x}{16} + C

Correct answerYour answer

\frac{\sin^{3} x\cos^{3} x}{9} + C

Correct answerYour answer

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AnswerA · $\frac{x}{8} - \frac{\sin 4x}{32} + C$

$\sin^{2} x\cos^{2} x = \frac{\sin^{2} 2x}{4} = \frac{1 - \cos 4x}{8}$ .
$\int \frac{1 - \cos 4x}{8}\,dx = \frac{x}{8} - \frac{\sin 4x}{32} + C$ .

Worth noting

A plus sign comes from $\cos^{2}$ , not $\sin^{2}$ : $\sin^{2} 2x = \frac{1 - \cos 4x}{2}$ .

Integrals of particular functions

15 questions

Question 46

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Evaluate $\int \frac{1}{x^{2} + 16}\,dx$ .

\tan^{-1}{\left(\frac{x}{4} \right)} + C

Correct answerYour answer

4 \tan^{-1}{\left(\frac{x}{4} \right)} + C

Correct answerYour answer

\frac{\tan^{-1}{\left(\frac{x}{4} \right)}}{4} + C

Correct answerYour answer

\log\left|x^{2} + 16\right| + C

Correct answerYour answer

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AnswerC · $\frac{\tan^{-1}{\left(\frac{x}{4} \right)}}{4} + C$

$\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$ .
Here $a = 4$ : $\frac{1}{4}\tan^{-1}\frac{x}{4} + C$ .

Worth noting

Without the factor $\frac{1}{4}$ the derivative comes out 4 times too large.

Question 47

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Evaluate $\int \frac{1}{\sqrt{25 - x^{2}}}\,dx$ .

\frac{\sin^{-1}{\left(\frac{x}{5} \right)}}{5} + C

Correct answerYour answer

\sin^{-1}{\left(\frac{x}{5} \right)} + C

Correct answerYour answer

\frac{\tan^{-1}{\left(\frac{x}{5} \right)}}{5} + C

Correct answerYour answer

\sqrt{25 - x^{2}} + C

Correct answerYour answer

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AnswerB · $\sin^{-1}{\left(\frac{x}{5} \right)} + C$

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$ .
With $a = 5$ : $\sin^{-1}\frac{x}{5} + C$ .

Worth noting

Unlike the $\tan^{-1}$ form, this one has no factor $\frac{1}{a}$ in front.

Question 48

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Evaluate $\int \frac{1}{x^{2} - 9}\,dx$ .

\frac{1}{6}\log\left|\frac{x + 3}{x - 3}\right| + C

Correct answerYour answer

\frac{1}{3}\tan^{-1}\frac{x}{3} + C

Correct answerYour answer

\log\left|x^2 - 9\right| + C

Correct answerYour answer

\frac{1}{6}\log\left|\frac{x - 3}{x + 3}\right| + C

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AnswerD · $\frac{1}{6}\log\left|\frac{x - 3}{x + 3}\right| + C$

$\frac{1}{x^2 - 9} = \frac{1}{6}\left(\frac{1}{x - 3} - \frac{1}{x + 3}\right)$ .
So the integral is $\frac{1}{6}\log\left|\frac{x - 3}{x + 3}\right| + C$ : the standard form $\frac{1}{2a}\log\left|\frac{x - a}{x + a}\right|$ .

Worth noting

The $\tan^{-1}$ form belongs to $x^2 + a^2$ , with a plus sign.

Question 49

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Evaluate $\int \frac{1}{\sqrt{x^{2} + 9}}\,dx$ .

\log\left|x + \sqrt{x^2 + 9}\right| + C

Correct answerYour answer

\sin^{-1}\frac{x}{3} + C

Correct answerYour answer

\frac{1}{3}\tan^{-1}\frac{x}{3} + C

Correct answerYour answer

2\sqrt{x^2 + 9} + C

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AnswerA · $\log\left|x + \sqrt{x^2 + 9}\right| + C$

$\int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C$ .
With $a = 3$ : $\log\left|x + \sqrt{x^2 + 9}\right| + C$ .

Worth noting

$\sin^{-1}\frac{x}{3}$ is for $\sqrt{9 - x^2}$ , with the minus sign inside.

Question 50

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Match each integrand in List I with its integral in List II ( $a > 0$ ; add $C$ to each).

List I	List II
A. $\frac{1}{x^2 + a^2}$	I. $\sin^{-1}\frac{x}{a}$
B. $\frac{1}{x^2 - a^2}$	II. $\frac{1}{a}\tan^{-1}\frac{x}{a}$
C. $\frac{1}{\sqrt{a^2 - x^2}}$	III. $\log\left\|x + \sqrt{x^2 + a^2}\right\|$
D. $\frac{1}{\sqrt{x^2 + a^2}}$	IV. $\frac{1}{2a}\log\left\|\frac{x - a}{x + a}\right\|$

A-IV, B-II, C-I, D-IIICorrect answerYour answerA-II, B-IV, C-III, D-ICorrect answerYour answerA-IV, B-II, C-III, D-ICorrect answerYour answerA-II, B-IV, C-I, D-IIICorrect answerYour answer

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AnswerD · A-II, B-IV, C-I, D-III

Each can be checked by differentiating: for example $\frac{d}{dx}\sin^{-1}\frac{x}{a} = \frac{1}{\sqrt{a^2 - x^2}}$ .
A plus sign without a root gives $\tan^{-1}$ ; a minus sign gives a logarithm of a ratio; the roots give $\sin^{-1}$ and the logarithm with $x + \sqrt{\;}$ .

Question 51

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Evaluate $\int \frac{1}{9 - x^{2}}\,dx$ .

\frac{1}{3}\log\left|\frac{3 + x}{3 - x}\right| + C

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\frac{1}{6}\log\left|\frac{3 - x}{3 + x}\right| + C

Correct answerYour answer

\frac{1}{6}\log\left|\frac{3 + x}{3 - x}\right| + C

Correct answerYour answer

\frac{1}{3}\tan^{-1}\frac{x}{3} + C

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AnswerC · $\frac{1}{6}\log\left|\frac{3 + x}{3 - x}\right| + C$

$\int \frac{dx}{a^{2} - x^{2}} = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + C$ .
With $a = 3$ : $\frac{1}{6}\log\left|\frac{3 + x}{3 - x}\right| + C$ .

Worth noting

$\frac{1}{3}$ uses $\frac{1}{a}$ where the formula has $\frac{1}{2a}$ .

Question 52

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Evaluate $\int \frac{1}{\sqrt{x^{2} - 16}}\,dx$ .

\log\left|x + \sqrt{x^{2} + 16}\right| + C

Correct answerYour answer

\log\left|x + \sqrt{x^{2} - 16}\right| + C

Correct answerYour answer

\frac{1}{4}\log\left|x + \sqrt{x^{2} - 16}\right| + C

Correct answerYour answer

\sqrt{x^{2} - 16} + C

Correct answerYour answer

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AnswerB · $\log\left|x + \sqrt{x^{2} - 16}\right| + C$

$\int \frac{dx}{\sqrt{x^{2} - a^{2}}} = \log\left|x + \sqrt{x^{2} - a^{2}}\right| + C$ .
With $a = 4$ : $\log\left|x + \sqrt{x^{2} - 16}\right| + C$ .

Worth noting

This formula has no $\frac{1}{a}$ in front, unlike $\int \frac{dx}{x^{2} + a^{2}}$ .

Question 53

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Evaluate $\int \frac{dx}{x^{2} + 2x + 2}$ .

\tan^{-1}(x - 1) + C

Correct answerYour answer

\frac{1}{2}\tan^{-1}\frac{x + 1}{2} + C

Correct answerYour answer

\log\left|x^{2} + 2x + 2\right| + C

Correct answerYour answer

\tan^{-1}(x + 1) + C

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AnswerD · $\tan^{-1}(x + 1) + C$

Complete the square: $x^{2} + 2x + 2 = (x + 1)^{2} + 1$ .
$\int \frac{dx}{(x + 1)^{2} + 1} = \tan^{-1}(x + 1) + C$ .

Worth noting

$\tan^{-1}(x - 1)$ gets the sign of the shift wrong: $x^{2} + 2x = (x + 1)^{2} - 1$ .

Question 54

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Evaluate $\int \frac{e^{x}}{e^{2x} + 1}\,dx$ .

\frac{1}{2}\log\left(e^{2x} + 1\right) + C

Correct answerYour answer

\tan^{-1}\left(e^{2x}\right) + C

Correct answerYour answer

\log\left(e^{x} + 1\right) + C

Correct answerYour answer

\tan^{-1}\left(e^{x}\right) + C

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AnswerD · $\tan^{-1}\left(e^{x}\right) + C$

Put $t = e^{x}$ , $dt = e^{x}\,dx$ : $\int \frac{dt}{t^{2} + 1}$ .
$= \tan^{-1} t = \tan^{-1}(e^{x}) + C$ .

Worth noting

$\frac{1}{2}\log(e^{2x} + 1)$ would need $e^{2x}$ on top, the derivative of the bottom over 2.

Question 55

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Read the two statements and choose the correct option.

I. $\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$ .

II. $\int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \frac{1}{a}\sin^{-1}\frac{x}{a} + C$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $\frac{d}{dx}\left(\frac{1}{a}\tan^{-1}\frac{x}{a}\right) = \frac{1}{a} \cdot \frac{1/a}{1 + x^{2}/a^{2}} = \frac{1}{x^{2} + a^{2}}$ . True.
II: $\frac{d}{dx}\sin^{-1}\frac{x}{a} = \frac{1}{\sqrt{a^{2} - x^{2}}}$ already, so the extra $\frac{1}{a}$ is wrong. False.

Question 56

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Evaluate $\int \frac{dx}{x^2 + 4x + 13}$ .

\tan^{-1}\frac{x + 2}{3} + C

Correct answerYour answer

\frac{1}{3}\tan^{-1}\frac{x - 2}{3} + C

Correct answerYour answer

\frac{1}{3}\tan^{-1}\frac{x + 2}{3} + C

Correct answerYour answer

\log\left|x^2 + 4x + 13\right| + C

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AnswerC · $\frac{1}{3}\tan^{-1}\frac{x + 2}{3} + C$

Complete the square: $x^2 + 4x + 13 = (x + 2)^2 + 9$ .
With $t = x + 2$ and $a = 3$ : $\int \frac{dt}{t^2 + 9} = \frac{1}{3}\tan^{-1}\frac{x + 2}{3} + C$ .

Worth noting

$\frac{x - 2}{3}$ gets the sign of the shift wrong: $x^2 + 4x$ is $(x + 2)^2 - 4$ .

Question 57

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Evaluate $\int \frac{1}{4 x^{2} + 9}\,dx$ .

\frac{1}{3}\tan^{-1}\frac{2x}{3} + C

Correct answerYour answer

\frac{1}{6}\tan^{-1}\frac{2x}{3} + C

Correct answerYour answer

\frac{1}{12}\tan^{-1}\frac{2x}{3} + C

Correct answerYour answer

\frac{1}{6}\tan^{-1}\frac{3x}{2} + C

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AnswerB · $\frac{1}{6}\tan^{-1}\frac{2x}{3} + C$

$4x^2 + 9 = (2x)^2 + 3^2$ . Put $t = 2x$ , so $dx = \frac{dt}{2}$ .
$\frac{1}{2}\int \frac{dt}{t^2 + 9} = \frac{1}{2} \cdot \frac{1}{3}\tan^{-1}\frac{t}{3} = \frac{1}{6}\tan^{-1}\frac{2x}{3} + C$ .

Worth noting

$\frac{1}{3}$ forgets the $\frac{1}{2}$ that $dx = \frac{dt}{2}$ brings.

Question 58

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Evaluate $\int \frac{dx}{\sqrt{3 + 2x - x^{2}}}$ .

\sin^{-1}\frac{x - 1}{2} + C

Correct answerYour answer

\sin^{-1}\frac{x + 1}{2} + C

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\frac{1}{2}\sin^{-1}\frac{x - 1}{2} + C

Correct answerYour answer

\sin^{-1}(x - 1) + C

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AnswerA · $\sin^{-1}\frac{x - 1}{2} + C$

Complete the square: $3 + 2x - x^{2} = 4 - (x - 1)^{2}$ .
$\int \frac{dx}{\sqrt{2^{2} - (x - 1)^{2}}} = \sin^{-1}\frac{x - 1}{2} + C$ .

Worth noting

$\frac{1}{2}$ in front belongs to $\int \frac{dx}{x^{2} + a^{2}}$ ; $\int \frac{dx}{\sqrt{a^{2} - u^{2}}} = \sin^{-1}\frac{u}{a}$ has no such factor.

Question 59

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Evaluate $\int \frac{x}{x^{4} + 1}\,dx$ .

\tan^{-1} x^{2} + C

Correct answerYour answer

\frac{1}{2}\tan^{-1} x^{2} + C

Correct answerYour answer

\frac{1}{4}\log\left(x^{4} + 1\right) + C

Correct answerYour answer

\frac{1}{2}\tan^{-1} x + C

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AnswerB · $\frac{1}{2}\tan^{-1} x^{2} + C$

Put $t = x^{2}$ , $dt = 2x\,dx$ : $\frac{1}{2}\int \frac{dt}{t^{2} + 1}$ .
$= \frac{1}{2}\tan^{-1} t = \frac{1}{2}\tan^{-1} x^{2} + C$ .

Worth noting

$\frac{1}{4}\log(x^{4} + 1)$ would need $x^{3}$ on top, the derivative of $x^{4}$ .

Question 60

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Evaluate $\int \frac{1}{\sqrt{9 x^{2} + 1}}\,dx$ .

\log\left|3x + \sqrt{9x^{2} + 1}\right| + C

Correct answerYour answer

\frac{1}{3}\tan^{-1} 3x + C

Correct answerYour answer

\frac{1}{3}\log\left|3x + \sqrt{9x^{2} + 1}\right| + C

Correct answerYour answer

\frac{1}{9}\log\left|3x + \sqrt{9x^{2} + 1}\right| + C

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AnswerC · $\frac{1}{3}\log\left|3x + \sqrt{9x^{2} + 1}\right| + C$

Put $t = 3x$ , $dt = 3\,dx$ : $\frac{1}{3}\int \frac{dt}{\sqrt{t^{2} + 1}}$ .
$= \frac{1}{3}\log\left|t + \sqrt{t^{2} + 1}\right| = \frac{1}{3}\log\left|3x + \sqrt{9x^{2} + 1}\right| + C$ .

Worth noting

Without the $\frac{1}{3}$ , the answer differentiates to three times the integrand.

Partial fractions

12 questions

Question 61

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If $\frac{5x + 4}{(x + 2)(x - 1)} = \frac{A}{x + 2} + \frac{B}{x - 1}$ , what is $A$ ?

-2

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1

Correct answerYour answer

2

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3

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AnswerC · $2$

Multiply through: $5x + 4 = A(x - 1) + B(x + 2)$ .
Put $x = -2$ : $-6 = -3A$ , so $A = 2$ . (And $x = 1$ gives $B = 3$ .)

Worth noting

3 is $B$ , the constant over $x - 1$ .

Question 62

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Which form of partial fractions fits $\frac{x + 2}{(x - 1)^{2}(x + 3)}$ ?

\frac{A}{(x - 1)^{2}} + \frac{B}{x + 3}

Correct answerYour answer

\frac{A}{x - 1} + \frac{B}{x + 3}

Correct answerYour answer

\frac{Ax + B}{(x - 1)^{2}} + \frac{C}{x + 3}

Correct answerYour answer

\frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{x + 3}

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AnswerD · $\frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{x + 3}$

A repeated factor $(x - 1)^{2}$ needs a term for each power: $\frac{A}{x - 1}$ and $\frac{B}{(x - 1)^{2}}$ .
The factor $x + 3$ gets its own $\frac{C}{x + 3}$ .

Worth noting

Keeping only $\frac{A}{(x - 1)^{2}}$ leaves too few constants to match every numerator.

Question 63

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If $\frac{x + 7}{(x - 1)(x + 3)} = \frac{A}{x - 1} + \frac{B}{x + 3}$ , what is $B$ ?

-2

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-1

Correct answerYour answer

1

Correct answerYour answer

2

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AnswerB · $-1$

$x + 7 = A(x + 3) + B(x - 1)$ .
Put $x = -3$ : $4 = -4B$ , so $B = -1$ . (And $x = 1$ gives $A = 2$ .)

Worth noting

2 is $A$ , the constant over $x - 1$ .

Question 64

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Evaluate $\int \frac{3 x + 1}{x^{2} - 1}\,dx$ .

2\log\left|x - 1\right| + \log\left|x + 1\right| + C

Correct answerYour answer

\log\left|x - 1\right| + 2\log\left|x + 1\right| + C

Correct answerYour answer

\frac{3}{2}\log\left|x^2 - 1\right| + C

Correct answerYour answer

2\log\left|x - 1\right| - \log\left|x + 1\right| + C

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AnswerA · $2\log\left|x - 1\right| + \log\left|x + 1\right| + C$

$\frac{3x + 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$ : $x = 1$ gives $4 = 2A$ , $x = -1$ gives $-2 = -2B$ , so $A = 2$ , $B = 1$ .
$\int \left(\frac{2}{x - 1} + \frac{1}{x + 1}\right)dx = 2\log|x - 1| + \log|x + 1| + C$ .

Worth noting

Swapping $A$ and $B$ puts the 2 on the wrong factor; put each root back in to check.

Question 65

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Evaluate $\int \frac{x}{\left(x - 2\right) \left(x - 1\right)}\,dx$ .

2\log\left|x - 1\right| - \log\left|x - 2\right| + C

Correct answerYour answer

2\log\left|x - 2\right| - \log\left|x - 1\right| + C

Correct answerYour answer

\log\left|(x - 1)(x - 2)\right| + C

Correct answerYour answer

\log\left|x - 1\right| + 2\log\left|x - 2\right| + C

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AnswerB · $2\log\left|x - 2\right| - \log\left|x - 1\right| + C$

$x = A(x - 2) + B(x - 1)$ : $x = 1$ gives $A = -1$ , $x = 2$ gives $B = 2$ .
$\int \left(-\frac{1}{x - 1} + \frac{2}{x - 2}\right)dx = 2\log|x - 2| - \log|x - 1| + C$ .

Worth noting

$\log|(x - 1)(x - 2)|$ would need both constants to be 1.

Question 66

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Evaluate $\int \frac{dx}{x^{2} - 5x + 6}$ .

\log\left|\frac{x - 2}{x - 3}\right| + C

Correct answerYour answer

\log\left|(x - 2)(x - 3)\right| + C

Correct answerYour answer

\log\left|\frac{x - 3}{x - 2}\right| + C

Correct answerYour answer

\frac{1}{2}\log\left|\frac{x - 3}{x - 2}\right| + C

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AnswerC · $\log\left|\frac{x - 3}{x - 2}\right| + C$

$\frac{1}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$ : $x = 2$ gives $A = -1$ , $x = 3$ gives $B = 1$ .
$\int \left(\frac{1}{x - 3} - \frac{1}{x - 2}\right)dx = \log\left|\frac{x - 3}{x - 2}\right| + C$ .

Worth noting

The fraction turned upside down comes from swapping the signs of $A$ and $B$ .

Question 67

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Evaluate $\int \frac{2x}{(x + 1)(x + 3)}\,dx$ .

3\log\left|x + 1\right| - \log\left|x + 3\right| + C

Correct answerYour answer

\log\left|(x + 1)(x + 3)\right| + C

Correct answerYour answer

\log\left|x + 1\right| + 3\log\left|x + 3\right| + C

Correct answerYour answer

3\log\left|x + 3\right| - \log\left|x + 1\right| + C

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AnswerD · $3\log\left|x + 3\right| - \log\left|x + 1\right| + C$

$2x = A(x + 3) + B(x + 1)$ : $x = -1$ gives $-2 = 2A$ , $A = -1$ ; $x = -3$ gives $-6 = -2B$ , $B = 3$ .
$\int \left(-\frac{1}{x + 1} + \frac{3}{x + 3}\right)dx = 3\log|x + 3| - \log|x + 1| + C$ .

Worth noting

Swapping $A$ and $B$ puts the 3 on the wrong factor; put each root back in to check.

Question 68

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Evaluate $\int \frac{x^{2} + 1}{x^{2} - 1}\,dx$ .

\log\left|\frac{x - 1}{x + 1}\right| + C

Correct answerYour answer

x + \log\left|\frac{x + 1}{x - 1}\right| + C

Correct answerYour answer

x + 2\log\left|x^2 - 1\right| + C

Correct answerYour answer

x + \log\left|\frac{x - 1}{x + 1}\right| + C

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AnswerD · $x + \log\left|\frac{x - 1}{x + 1}\right| + C$

The fraction is improper (equal degrees), so divide first: $\frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1}$ .
$\frac{2}{x^2 - 1} = \frac{1}{x - 1} - \frac{1}{x + 1}$ , so the integral is $x + \log\left|\frac{x - 1}{x + 1}\right| + C$ .

Worth noting

Leaving out the $x$ forgets the 1 that the division gives.

Question 69

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Evaluate $\int \frac{1}{x \left(x + 1\right)^{2}}\,dx$ .

\log\left|\frac{x}{x + 1}\right| + \frac{1}{x + 1} + C

Correct answerYour answer

\log\left|\frac{x}{x + 1}\right| - \frac{1}{x + 1} + C

Correct answerYour answer

\log\left|\frac{x}{x + 1}\right| + C

Correct answerYour answer

\log\left|x\right| - \frac{1}{x + 1} + C

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AnswerA · $\log\left|\frac{x}{x + 1}\right| + \frac{1}{x + 1} + C$

A repeated factor needs two fractions: $\frac{1}{x(x + 1)^2} = \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{(x + 1)^2}$ .
$\int -\frac{dx}{(x + 1)^2} = \frac{1}{x + 1}$ , so the integral is $\log\left|\frac{x}{x + 1}\right| + \frac{1}{x + 1} + C$ .

Worth noting

$-\frac{1}{x + 1}$ gets the sign of $\int -\frac{dx}{(x + 1)^2}$ wrong: integrating $(x + 1)^{-2}$ gives $-(x + 1)^{-1}$ .

Question 70

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Evaluate $\int \frac{dx}{x(x^{2} + 1)}$ .

\log\left|x\right| + \frac{1}{2}\log\left(x^{2} + 1\right) + C

Correct answerYour answer

\log\left|x\right| - \log\left(x^{2} + 1\right) + C

Correct answerYour answer

\log\left|x\right| - \frac{1}{2}\log\left(x^{2} + 1\right) + C

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\log\left|x\right| - \tan^{-1} x + C

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AnswerC · $\log\left|x\right| - \frac{1}{2}\log\left(x^{2} + 1\right) + C$

$\frac{1}{x(x^{2} + 1)} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}$ gives $A = 1$ , $B = -1$ , $C = 0$ .
$\int \left(\frac{1}{x} - \frac{x}{x^{2} + 1}\right)dx = \log|x| - \frac{1}{2}\log(x^{2} + 1) + C$ .

Worth noting

$\int \frac{x}{x^{2} + 1}\,dx = \frac{1}{2}\log(x^{2} + 1)$ : the $\frac{1}{2}$ comes from $d(x^{2} + 1) = 2x\,dx$ .

Question 71

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Evaluate $\int \frac{x^{2}}{x^{2} - 4}\,dx$ .

x + \log\left|\frac{x - 2}{x + 2}\right| + C

Correct answerYour answer

\log\left|\frac{x - 2}{x + 2}\right| + C

Correct answerYour answer

x + \log\left|\frac{x + 2}{x - 2}\right| + C

Correct answerYour answer

x + 4\log\left|\frac{x - 2}{x + 2}\right| + C

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AnswerA · $x + \log\left|\frac{x - 2}{x + 2}\right| + C$

The fraction is improper: $\frac{x^{2}}{x^{2} - 4} = 1 + \frac{4}{x^{2} - 4}$ .
$\int \frac{4\,dx}{x^{2} - 4} = 4 \cdot \frac{1}{4}\log\left|\frac{x - 2}{x + 2}\right|$ , so the integral is $x + \log\left|\frac{x - 2}{x + 2}\right| + C$ .

Worth noting

Leaving out $x$ forgets the 1 that the division gives first.

Question 72

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Evaluate $\int \frac{3x - 1}{(x - 2)^{2}}\,dx$ .

3\log\left|x - 2\right| + \frac{5}{x - 2} + C

Correct answerYour answer

3\log\left|x - 2\right| - \frac{5}{x - 2} + C

Correct answerYour answer

8\log\left|x - 2\right| + C

Correct answerYour answer

3\log\left|x - 2\right| - \frac{1}{x - 2} + C

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AnswerB · $3\log\left|x - 2\right| - \frac{5}{x - 2} + C$

$\frac{3x - 1}{(x - 2)^{2}} = \frac{A}{x - 2} + \frac{B}{(x - 2)^{2}}$ : $3x - 1 = A(x - 2) + B$ , so $A = 3$ and, at $x = 2$ , $B = 5$ .
$\int \frac{5}{(x - 2)^{2}}\,dx = -\frac{5}{x - 2}$ , so the integral is $3\log|x - 2| - \frac{5}{x - 2} + C$ .

Worth noting

$+\frac{5}{x - 2}$ loses the sign: $\int (x - 2)^{-2}\,dx = -(x - 2)^{-1}$ .

Integration by parts

18 questions

Question 73

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Evaluate $\int x e^{x}\,dx$ .

\left(x + 1\right) e^{x} + C

Correct answerYour answer

x e^{x} + C

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\left(x - 1\right) e^{x} + C

Correct answerYour answer

- \left(x - 1\right) e^{x} + C

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AnswerC · $\left(x - 1\right) e^{x} + C$

By parts, $\int u\,dv = uv - \int v\,du$ , with $u = x$ and $dv = e^{x}\,dx$ , so $v = e^{x}$ .
$x e^{x} - \int e^{x}\,dx = \left(x - 1\right) e^{x} + C$ .

Worth noting

Adding the second integral instead of subtracting it is the slip one of the options is built on.

Question 74

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To evaluate $\int x^2\log x\,dx$ by parts, which should be taken as $u$ , the function to differentiate?

\log x

Correct answerYour answer

x^2

Correct answerYour answer

x^2\log x

Correct answerYour answerEither choice works equally wellCorrect answerYour answer

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AnswerA · $\log x$

We cannot integrate $\log x$ easily, but we can differentiate it to $\frac{1}{x}$ , which cancels a power of $x$ .
So $u = \log x$ and $dv = x^2\,dx$ . The order Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential (ILATE) puts the logarithm before the algebraic factor.

Worth noting

With $u = x^2$ , $dv = \log x\,dx$ has to be integrated first, which is itself a by-parts problem.

Question 75

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Evaluate $\int x \cos x\,dx$ .

x \sin x - \cos x + C

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x \sin x + \cos x + C

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x \sin x + C

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- x \sin x - \cos x + C

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AnswerB · $x \sin x + \cos x + C$

By parts, $\int u\,dv = uv - \int v\,du$ , with $u = x$ and $dv = \cos x\,dx$ , so $v = \sin x$ .
$x \sin x - \int \sin x\,dx = x \sin x + \cos x + C$ .

Worth noting

Adding the second integral instead of subtracting it is the slip one of the options is built on.

Question 76

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To evaluate $\int x\tan^{-1} x\,dx$ by parts, which should be taken as $u$ , the function to differentiate?

x

Correct answerYour answer

x\tan^{-1} x

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1

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\tan^{-1} x

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AnswerD · $\tan^{-1} x$

The order of preference for $u$ is inverse trigonometric, then logarithmic, algebraic, trigonometric, exponential.
$\tan^{-1} x$ comes first: its derivative $\frac{1}{1 + x^{2}}$ is simple, while $\int \tan^{-1} x\,dx$ is not.

Worth noting

Taking $u = x$ leaves $dv = \tan^{-1} x\,dx$ , which we cannot integrate directly.

Question 77

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Evaluate $\int e^x\left(\tan x + \sec^2 x\right)dx$ .

e^x \sec^2 x + C

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e^x\left(\tan x + \sec^2 x\right) + C

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e^{x} \tan x + C

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- e^{x} \tan x + C

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AnswerC · $e^{x} \tan x + C$

$\int e^x\left(f(x) + f'(x)\right)dx = e^x f(x) + C$ , because $\frac{d}{dx}\left(e^x f(x)\right) = e^x\left(f(x) + f'(x)\right)$ .
Here $f(x) = \tan x$ and $f'(x) = \sec^2 x$ , so the integral is $e^{x} \tan x + C$ .

Worth noting

The function that goes with $e^x$ is the one whose derivative is the other term.

Question 78

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Evaluate $\int \log x\,dx$ .

x \left(\log x + 1\right) + C

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x \log x + C

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- x \left(\log x - 1\right) + C

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x \left(\log x - 1\right) + C

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AnswerD · $x \left(\log x - 1\right) + C$

By parts, $\int u\,dv = uv - \int v\,du$ , with $u = \log x$ and $dv = 1\,dx$ , so $v = x$ .
$x \log x - \int 1\,dx = x \left(\log x - 1\right) + C$ .

Worth noting

Adding the second integral instead of subtracting it is the slip one of the options is built on.

Question 79

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Evaluate $\int e^x\left(x^{2} + 2x\right)dx$ .

e^x 2x + C

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x^{2} e^{x} + C

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e^x\left(x^{2} + 2x\right) + C

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- x^{2} e^{x} + C

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AnswerB · $x^{2} e^{x} + C$

$\int e^x\left(f(x) + f'(x)\right)dx = e^x f(x) + C$ , because $\frac{d}{dx}\left(e^x f(x)\right) = e^x\left(f(x) + f'(x)\right)$ .
Here $f(x) = x^{2}$ and $f'(x) = 2x$ , so the integral is $x^{2} e^{x} + C$ .

Worth noting

The function that goes with $e^x$ is the one whose derivative is the other term.

Question 80

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Evaluate $\int x \log x\,dx$ .

\frac{x^{2} \left(2 \log x - 1\right)}{4} + C

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\frac{x^{2} \left(2 \log x + 1\right)}{4} + C

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\frac{x^{2} \log x}{2} + C

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- \frac{x^{2} \left(2 \log x - 1\right)}{4} + C

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AnswerA · $\frac{x^{2} \left(2 \log x - 1\right)}{4} + C$

By parts, $\int u\,dv = uv - \int v\,du$ , with $u = \log x$ and $dv = x\,dx$ , so $v = \frac{x^{2}}{2}$ .
$\frac{x^{2} \log x}{2} - \int \frac{x}{2}\,dx = \frac{x^{2} \left(2 \log x - 1\right)}{4} + C$ .

Worth noting

Adding the second integral instead of subtracting it is the slip one of the options is built on.

Question 81

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Evaluate $\int e^{x}\left(\frac{1}{x} - \frac{1}{x^{2}}\right)dx$ .

-\frac{e^{x}}{x^{2}} + C

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e^{x}\log x + C

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\frac{e^{x}}{x^{2}} + C

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\frac{e^{x}}{x} + C

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AnswerD · $\frac{e^{x}}{x} + C$

This is $e^{x}\left(f(x) + f'(x)\right)$ with $f(x) = \frac{1}{x}$ , since $f'(x) = -\frac{1}{x^{2}}$ .
$\int e^{x}(f + f')\,dx = e^{x}f(x) + C = \frac{e^{x}}{x} + C$ .

Worth noting

$-\frac{e^{x}}{x^{2}}$ picks $f'$ as the function; it must be $f$ , the part whose derivative is also there.

Question 82

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Evaluate $\int x\sin 2x\,dx$ .

-\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C

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-\frac{x\cos 2x}{2} - \frac{\sin 2x}{4} + C

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\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C

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-x\cos 2x + \frac{\sin 2x}{2} + C

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AnswerA · $-\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C$

By parts with $u = x$ , $dv = \sin 2x\,dx$ : $v = -\frac{\cos 2x}{2}$ .
$-\frac{x\cos 2x}{2} + \int \frac{\cos 2x}{2}\,dx = -\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C$ .

Worth noting

$-\frac{\sin 2x}{4}$ slips on the double minus: $-\int v\,du = +\int \frac{\cos 2x}{2}\,dx$ .

Question 83

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Evaluate $\int x^{2}\log x\,dx$ .

\frac{x^{3}\log x}{3} + \frac{x^{3}}{9} + C

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\frac{x^{3}\log x}{3} - \frac{x^{3}}{3} + C

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\frac{x^{3}\log x}{3} - \frac{x^{3}}{9} + C

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2x\log x + x + C

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AnswerC · $\frac{x^{3}\log x}{3} - \frac{x^{3}}{9} + C$

Take $u = \log x$ and $dv = x^{2}\,dx$ : $du = \frac{dx}{x}$ , $v = \frac{x^{3}}{3}$ .
$\frac{x^{3}\log x}{3} - \int \frac{x^{2}}{3}\,dx = \frac{x^{3}\log x}{3} - \frac{x^{3}}{9} + C$ .

Worth noting

$-\frac{x^{3}}{3}$ forgets to integrate $\frac{x^{2}}{3}$ , which gives $\frac{x^{3}}{9}$ .

Question 84

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Evaluate $\int xe^{2x}\,dx$ .

\frac{xe^{2x}}{2} + \frac{e^{2x}}{4} + C

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\frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C

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xe^{2x} - \frac{e^{2x}}{2} + C

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\frac{x^{2}e^{2x}}{4} + C

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AnswerB · $\frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C$

$u = x$ , $dv = e^{2x}\,dx$ : $v = \frac{e^{2x}}{2}$ .
$\frac{xe^{2x}}{2} - \int \frac{e^{2x}}{2}\,dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C$ .

Worth noting

$+\frac{e^{2x}}{4}$ adds what by parts subtracts: $\int u\,dv = uv - \int v\,du$ .

Question 85

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Evaluate $\int \tan^{-1}{\left(x \right)}\,dx$ .

x\tan^{-1} x + \frac{1}{2}\log(1 + x^2) + C

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\frac{1}{1 + x^2} + C

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x\tan^{-1} x - \log(1 + x^2) + C

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x\tan^{-1} x - \frac{1}{2}\log(1 + x^2) + C

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AnswerD · $x\tan^{-1} x - \frac{1}{2}\log(1 + x^2) + C$

Take $u = \tan^{-1} x$ and $dv = dx$ , the hidden factor 1: then $v = x$ .
$x\tan^{-1} x - \int \frac{x}{1 + x^2}\,dx = x\tan^{-1} x - \frac{1}{2}\log(1 + x^2) + C$ .

Worth noting

$\int \frac{x}{1 + x^2}\,dx$ is $\frac{1}{2}\log(1 + x^2)$ ; dropping the $\frac{1}{2}$ gives $x\tan^{-1} x - \log(1 + x^2)$ .

Question 86

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Evaluate $\int x^{2} e^{x}\,dx$ .

\left(x^{2} + 2 x - 2\right) e^{x} + C

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x^{2} e^{x} + C

Correct answerYour answer

\left(x^{2} - 2 x + 2\right) e^{x} + C

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- \left(x^{2} - 2 x + 2\right) e^{x} + C

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AnswerC · $\left(x^{2} - 2 x + 2\right) e^{x} + C$

By parts, $\int u\,dv = uv - \int v\,du$ , with $u = x^{2}$ and $dv = e^{x}\,dx$ , so $v = e^{x}$ .
$x^{2} e^{x} - \int 2 x e^{x}\,dx = \left(x^{2} - 2 x + 2\right) e^{x} + C$ .

Worth noting

Adding the second integral instead of subtracting it is the slip one of the options is built on.

Question 87

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Evaluate $\int x \sec^{2} x\,dx$ .

x\tan x + \log\left|\sec x\right| + C

Correct answerYour answer

x\tan x - \log\left|\sec x\right| + C

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x \tan x + C

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-\left(x\tan x - \log\left|\sec x\right|\right) + C

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AnswerB · $x\tan x - \log\left|\sec x\right| + C$

By parts, $\int u\,dv = uv - \int v\,du$ , with $u = x$ and $dv = \sec^{2} x\,dx$ , so $v = \frac{\sin x}{\cos x}$ .
$\frac{x \sin x}{\cos x} - \int \tan x\,dx = x\tan x - \log\left|\sec x\right| + C$ .

Worth noting

Adding the second integral instead of subtracting it is the slip one of the options is built on.

Question 88

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Evaluate $\int e^{x}\sin x\,dx$ .

\frac{e^{x}(\sin x - \cos x)}{2} + C

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\frac{e^{x}(\sin x + \cos x)}{2} + C

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e^{x}(\sin x - \cos x) + C

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-e^{x}\cos x + C

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AnswerA · $\frac{e^{x}(\sin x - \cos x)}{2} + C$

By parts twice: $I = e^{x}\sin x - \int e^{x}\cos x\,dx = e^{x}\sin x - e^{x}\cos x - I$ .
So $2I = e^{x}(\sin x - \cos x)$ and $I = \frac{e^{x}(\sin x - \cos x)}{2} + C$ .

Worth noting

$\frac{e^{x}(\sin x + \cos x)}{2}$ differentiates to $e^{x}\cos x$ , the other integral of the pair.

Question 89

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Evaluate $\int \sin^{-1} x\,dx$ .

x\sin^{-1} x - \sqrt{1 - x^{2}} + C

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\frac{1}{\sqrt{1 - x^{2}}} + C

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x\sin^{-1} x + \sqrt{1 - x^{2}} + C

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x\sin^{-1} x + C

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AnswerC · $x\sin^{-1} x + \sqrt{1 - x^{2}} + C$

Write it as $\int 1 \cdot \sin^{-1} x\,dx$ with $u = \sin^{-1} x$ , $dv = dx$ : $x\sin^{-1} x - \int \frac{x}{\sqrt{1 - x^{2}}}\,dx$ .
$\int \frac{x}{\sqrt{1 - x^{2}}}\,dx = -\sqrt{1 - x^{2}}$ , so the answer is $x\sin^{-1} x + \sqrt{1 - x^{2}} + C$ .

Worth noting

$-\sqrt{1 - x^{2}}$ loses the sign: subtracting $-\sqrt{1 - x^{2}}$ adds it.

Question 90

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Evaluate $\int \frac{\log x}{x^{2}}\,dx$ .

-\frac{\log x}{x} + \frac{1}{x} + C

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-\frac{\log x}{x} - \frac{1}{x} + C

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\frac{\log x}{x} - \frac{1}{x} + C

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\frac{(\log x)^{2}}{2} + C

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AnswerB · $-\frac{\log x}{x} - \frac{1}{x} + C$

Take $u = \log x$ , $dv = x^{-2}\,dx$ : $v = -\frac{1}{x}$ .
$-\frac{\log x}{x} + \int \frac{1}{x^{2}}\,dx = -\frac{\log x}{x} - \frac{1}{x} + C$ .

Worth noting

$+\frac{1}{x}$ treats $\int x^{-2}\,dx$ as $+\frac{1}{x}$ ; it is $-\frac{1}{x}$ .

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.