Question

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

Answer 1

$\int \frac{{\tan }^{-1}xdx}{1+x^2}$

Let $t={\tan }^{-1}x\Rightarrow dt=\frac{dx}{1+x^2}$

$=\int t.dt$

$=\frac{t^2}{2}+c$ where $c$ is the constant of integration.

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