Question

Simplify:
$\int x{\tan }^{-1}xdx$

Answer 1

Consider the given integral.

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

We know that

$\int uvdx=u\int vdx-\int (\frac{d}{dx}\left(u\right)\int vdx)dx$

Therefore,

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

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

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

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

$I=\frac{x^2}{2}{\tan }^{-1}x-\frac{1}{2}[x-{\tan }^{-1}x]+C$

Hence, this is the answer.