Evaluate: ∫ x tan⁻¹ x dx

Integrate by parts with $u=\tan^{-1}x$, $v=x$: $\int x\tan^{-1}x\,dx=\frac{x^2}{2}\tan^{-1}x-\frac12\int\frac{x^2}{1+x^2}dx$ $=\frac{x^2}{2}\tan^{-1}x-\frac12\int\left(1-\frac{1}{1+x^2}\right)dx$ $=\frac{x^2}{2}\tan^{-1}x-\frac12\left(x-\tan^{-1}x\right)+C$ Hence the integral is $\frac{x^2+1}{2}\tan^{-1}x-\frac{x}{2}+C$.