# Evaluate : $\int x\: tan^{-1}x\: dx$

Toolbox:
• When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts $\int udv=uv-\int vdu$
• $\;\large\frac{d}{dx}$$(\tan^{-1}x)=\large\frac{1}{1+x^2}. • Method of substitution :If f(x)=t,then f'(x)dx=dt. • Hence \int f(x)=\int t.dt+c. Step 1: I=\int x\tan^{-1}xdx This is of the form \int udv.This can be solved by uv-\int vdu Let u=\tan^{-1}x \Rightarrow du=\large\frac{1}{1+x^2}$$dx$
$v=\large\frac{x^2}{2}$
$dv=xdx$
Step 2:
$\therefore \int x\tan^{-1}xdx=(\tan^{-1}x)(\large\frac{x^2}{2})-\int \large\frac{x^2}{2}\frac{1}{1+x^2}$$dx \qquad\qquad\qquad=\large\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int \frac{x^2}{1+x^2}$$dx$
$\qquad\qquad\qquad=\large\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int \frac{x^2+1-1}{1+x^2}$$dx \qquad\qquad\qquad=\large\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int$$ dx-\int \large\frac{1}{1+x^2}$$dx \qquad\qquad\qquad=\large\frac{x^2\tan^{-1}x}{2}-\frac{x}{2}+\large\frac{\tan^{-1}x}{2}$$+c$