$\displaystyle \int x\tan^{-1}(x) dx$

let u = tan^-1(x)

du = $\displaystyle \frac{1}{1+x^2}dx$

let v = $\displaystyle \frac{x^2}{2}$

dv = $\displaystyle x$

then got stuck after this

$\displaystyle \frac{x^2}{2}\tan^{-1}{x} - \int{\frac{x^2}{2}(\frac{1}{1+x^2})}dx$