Question

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

Answer 1

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

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

$\Rightarrow \left(\frac{1}{1+x^2}\right)dx=dt$

Therefore,

$I=\int t.dt=\frac{t^2}{2}+c$

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

Hence, this is the answer.