Evaluate $\int_0^R \frac{r^2}{(1+r^2)^2}dr.$ I am trying to evaluate the following integral: $$\int_0^R \frac{r^2}{(1+r^2)^2}dr.$$ I might substitute $u=r^2$, but I don't find $du$ anywhere. Obviously the integral should be bounded on $R\in [0,\infty)$.
Any ideas?

Substitute $r=\tan{t}$; $dr = \sec^2{t} \, dt$.  Then the integral is equal to
$$\int_0^{\arctan{R}} dt \, \sec^2{t} \, \frac{\tan^2{t}}{ \sec^4{t}} = \int_0^{\arctan{R}} dt \,\sin^2{t}$$
which is then straightforward to evaluate:
$$\frac12 \left [t - \sin{t} \cos{t} \right ]_{0}^{\arctan{R}} = \frac12 \left [\arctan{R} - \frac{R}{1+R^2}\right ]$$
