Question

Integrate with respect o $x$:
$\frac{x^2{\tan }^{-1}x}{1+x^2}$

Answer 1

$\int \frac{x^2\left({\text{tan}}^{-1}x\right)}{1+x^2}dx=\int {\text{tan}}^{-1}x(1-\frac{1}{1+x^2})dx=\int ({\text{tan}}^{-1}x-(\frac{{\text{tan}}^{-1}x}{1+x^2}))dx$

$=x{\text{tan}}^{-1}x-\frac{1}{2}\ln (1+x^2)-\frac{1}{2}({\text{tan}}^{-1}x{)}^2$