Integrating $\frac{1}{(x^2+1)^2}$ How do I integrate $\frac{1}{(x^2+1)^2}$?
I've tried to use the fact that $\int \frac{1}{(x^2+1)}=Arctan(x)$ but I don't know how to. I think it's by parts. Tried using $u'=(x^2+1)^{-1}$ and $v=(x^2+1)^{-1}$ in the formula $\int u'v=vu-\int uv'$, but it doesn't seem to help...

First consider
$$I = \int \frac{dx}{(x^2+1)^{2}}$$
and the substitution $x=\tan\theta$. This leads to $dx = sec^{2}\theta \, d\theta$ and $x^2 + 1 = sec^{2}\theta$ and
\begin{align}
I &= \int \frac{sec^{2}\theta}{sec^{4}\theta} \, d\theta \\
&= \int \frac{d\theta}{sec^{2}\theta} \\
&= \int \cos^{2}\theta \, d\theta \\
&= \frac{1}{2} \, \int \left( 1 + \cos(2\theta) \right) \, d\theta\\
&= \frac{1}{2} \, \left[ \theta + \frac{1}{2} \, \sin(2\theta) \right] + c_{0}
\end{align}
Upon backward substitution this becomes
\begin{align}
I = \frac{1}{2} \, \tan^{-1}(x) + \frac{1}{4} \, \sin(2 \, \tan^{-1}(x)) + c_{0}
\end{align}
