Evaluate ∫ 1/(4 x^2 + 9) dx.

Question 57

HardOriginal question

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

$4x^2 + 9 = (2x)^2 + 3^2$ . 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.

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