Question

Integrate:

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

Answer 1

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

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

$\Rightarrow \int tdt$

By using $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$

$=\frac{t^2}{2}+c$

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