Evaluate ∫ 1/(x (x + 1)^2) dx.

Question 69

HardOriginal question

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

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 $-(x + 1)^{-1}$ .

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