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?
 A: 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:
$$\frac12 \left [t - \sin{t} \cos{t} \right ]_{0}^{\arctan{R}} = \frac12 \left [\arctan{R} - \frac{R}{1+R^2}\right ]$$
A: Denominator containing $1+r^2,$ let us set $r=\tan x$. Then
$$\begin{multline}
\int_0^R\frac{r^2}{(1+r^2)^2}\,dr
=\int_0^{\arctan R}\frac{\tan^2x}{\sec^4x}\sec^2x\,dx
=\int_0^{\arctan R}\sin^2x\,dx\\
=\int_0^{\arctan R}\frac{(1-\cos2x)}2\,dx
=\frac12\left(x-\frac {\sin2x}2\right)_0^{\arctan R} \\
=\frac12\left(\arctan R-\frac R{1+R^2} \right)
\end{multline}$$
as $$\sin(\arctan R)=2\frac{\tan(\arctan R)}{1+\tan^2(\arctan R)}=\frac{2R}{1+R^2}$$

Alternatively, integrating by parts,
$$\begin{multline}
\int\frac{r^2}{(1+r^2)^2}\,dr
=\int r\cdot \frac r{(1+r^2)^2}\,dr \\
=r\cdot \frac{-1}{2(1+r^2)}-\int \left(\frac{dr}{dr}\cdot \frac{-1}{2(1+r^2)}\right)\,dx
\end{multline}$$
$$=\frac{-r}{2(1+r^2)}+\frac12\int\frac{dr}{1+r^2}=-\frac r{2(1+r^2)}+\frac12\arctan r$$
A: Hint:$$\int_0^R \frac{r^2}{(1+r^2)^2}dr=\int_0^R \frac{r^2+1-1}{(1+r^2)^2}dr=\color{red}{\int_0^R \frac{1}{(1+r^2)}dr}-\color{green}{\int_0^R \frac{1}{(1+r^2)^2}dr}$$
answer of red integral is obvious .
for green integral substitute $r=\sin hx\to dr=\cos hx$ and $$\color{green}{\int_0^R \frac{1}{(1+r^2)^2}dr=\int_0^R \frac{\ coshx}{(1+\sinh x^2)^2}=\int_0^R \frac{dx}{cos^3hx}=\int_0^R \frac{dx}{\ coshxcos^2hx}=\int_0^R\ sechx(1-\tan^2h x)}=....... $$
