Question

Integrate the function $x{\tan }^{-1}x$

Answer 1

Let $I=\int x{\tan }^{-1}xdx$
Taking ${\tan }^{-1}x$ as first function and $x$ as second function and integrating by parts, we obtain
$I={\tan }^{-1}x\int xdx-\int \{\left(\frac{d}{dx}ta{n}^{-1}x\right)\int xdx\}dx$
$={\tan }^{-1}x\left(\frac{x^2}{2}\right)-\int \frac{1}{1+x^2}\cdot \frac{x^2}{2}dx$
$=\frac{x^2{\tan }^{-1}x}{2}-\frac{1}{2}\int \frac{x^2}{1+x^2}dx$
$=\frac{x^2{\tan }^{-1}x}{2}-\frac{1}{2}\int (\frac{x^2+1}{1+x^2}-\frac{1}{1+x^2})dx$
$=\frac{x^2{\tan }^{-1}x}{2}-\frac{1}{2}\int (1-\frac{1}{1+x^2})dx$
$=\frac{x^2{\tan }^{-1}x}{2}-\frac{1}{2}(x-{\tan }^{-1}x)+C$
$=\frac{x}{2}{\tan }^{-1}x-\frac{x}{2}+\frac{1}{2}{\tan }^{-1}x+C$