Evaluate ∫ tan^-1(x ) dx.

xtan^-1 x - 1/2log(1 + x^2) + C

xtan^-1 x + 1/2log(1 + x^2) + C

Question 85

HardOriginal question

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

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

Correct answerYour answer

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

Correct answerYour answer

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

Correct answerYour answer

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

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

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