Question

integrate :
$\int {\tan }^{-1}xdx.$

Answer 1

$I=\int {\tan }^{-1}xdx=\int 1.{\tan }^{-1}xdx$

Integration by parts

$I=x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx=x{\tan }^{-1}x-\frac{1}{2}\int \frac{2x}{1+x^2}dx=x{\tan }^{-1}x-\frac{1}{2}\left(\frac{n}{1+x^2}\right)+C$