Question

Evaluate :
$\int x^2{\tan }^{-1}\frac{x}{2}dx$

Answer 1

$I=\int \underset{II}{x^2}\underset{I}{{\tan }^{-1}}\left(\frac{3}{2}\right)dx$

$I=\left[\frac{x^3}{3}\right]{\tan }^{-1}\left(\frac{x}{2}\right)-\int \frac{1}{1+\frac{x^2}{4}}\times \frac{x^3}{3}dx$

$I=\frac{x^3}{3}{\tan }^{-1}\left(\frac{x}{2}\right)-\frac{1}{6}\int \frac{4x^3}{x^2+4}dx$

$I=\frac{x^3}{3}{\tan }^{-1}\left(\frac{x}{2}\right)-\frac{1}{6}\int (4x-\frac{16x}{x^2+4})dx$

$I=\frac{x^3}{3}{\tan }^{-1}\left(\frac{x}{2}\right)-\frac{1}{6}[\frac{4x^2}{2}-\frac{16}{2}log(x^2+4\left)\right]+c$

$I=\frac{x^3}{3}{\tan }^{-1}\left(\frac{x}{2}\right)-\frac{x^2}{3}+\frac{4}{3}log(x^2+4)+c$.