Question

Evaluate : $\int \frac{{\tan }^{-1}x}{1+x^2}dx$

Answer 1

We have,

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

Let $t={\tan }^{-1}x$

$dt=\frac{1}{1+x^2}dx$

Therefore,

$I=\int tdt$

$I=\frac{t^2}{2}+C$

$I=\frac{({\tan }^{-1}x{)}^2}{2}+C$

Hence, this is the answer.