Question

The value of $\int {\tan }^{-1}x$ dx is equal to

Answer 1

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

$\int v.dv$

$=uv-\int v.dv$

Let $v={\tan }^{-1}x;dv=1dx$

$dv=\frac{1}{x^2+1}dx$ and $v=x$

$I={\tan }^{-1}\left(x\right)x-\frac{1}{2}\int \frac{1}{v}dv$

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

As $v=x^2+1$

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