Question

Evaluate
$\int x^2{\tan }^{-1}xdx$

Answer 1

$\begin{array}{c}\int x^2{\tan }^{-1}dx=\frac{x^3}{3}{\tan }^{-1}x-\int \frac{1}{1+x^2}\times \frac{x^3}{3}dx\\ {\tan }^{-1}x\left(\frac{x^3}{3}\right)-\frac{1}{3}\int \frac{x^3}{1+x^2}dx\\ 1+x^2=t2dx=dt\\ dx=\frac{dt}{2x}\\ {\tan }^{-1}\left(\frac{x^3}{3}\right)-\frac{1}{3}\int \frac{x^3}{t}dt\\ {\tan }^{-1}\left(\frac{x^3}{3}\right)-\frac{1}{6}\left(1+x^2-\log \left|1+x^2\right|+c\right)\end{array}$