Question

Find the value of: $\int x^2{\tan }^{-1}xdx$.

Answer 1

$I=\int x^{2}ta{n}^{-1}xdx$

$II$ $I$

By ILETS Rule

$I=ta{n}^{-1}x\left[\frac{x^3}{3}\right]-\int \frac{1}{1+x^2}\times \frac{x^3}{3}dx$

$=\frac{x^3}{3}ta{n}^{-1}-\frac{1}{3}\int \frac{x^3}{1+x^2}dx$

Let $x^2=t$

$2xdx=dt$

$=\frac{x^3}{3}ta{n}^{-1}x-\frac{1}{3}\int \frac{tdt}{2(1+t)}$

$=\frac{x^3}{3}ta{n}^{-1}x-\frac{1}{6}\int (\frac{t+1}{t+1}-\frac{1}{t+1})dt$

$=\frac{x^3}{3}ta{n}^{-1}x-\frac{1}{6}\int (1-\frac{1}{1+t})dt$

$=\frac{x^3}{3}ta{n}^{-1}x-\frac{1}{6}[t-ln(1+t\left)\right]+c$

$=\frac{x^3}{3}ta{n}^{-1}x-\frac{1}{6}[x^2-ln(1+x^2\left)\right]+c$