$\displaystyle \int x tan^- (x) dx$ ( The Tan is inverse, i'm not sure how to properly display that in latex )

i did

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

Is it right ? If yes

now how can I solve $\displaystyle \int (\frac{x^2}{2}) (\frac{1}{1+x}) dx$