Evaluate the integral: $ \int x \tan^{-1}\ x \,\mathrm{d}x$

Notice that $$\frac{x^2}{x^2+1} = \frac{x^2+1-1}{x^2+1} = 1-\frac{1}{1+x^2}$$

Make polynomial division, then use $\left(\arctan(x)\right)'=\frac{1}{1+x^2}$.

Write it as $1-\dfrac1{1+x^2}$ and integrate term-wise.