Calculating $\int \frac{dx}{(x^2+1)^3}$ integral Would anyone help me calculate the following integral? $\int \frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $\int \frac{dx}{(x^2+1)^2}$ like that: 
$\int \frac{dx}{(x^2+1)^2} = \int \frac{x^2+1-x^2}{(x^2+1)^2}dx = \int \frac{1}{x^2+1}dx -  \int \frac{x^2}{(x^2+1)^2}dx = $
$= \Biggr\rvert \begin{equation}
\begin{split}
& u = x \quad v' =\frac{x}{(x^2+1)^2} =\frac{1(x^2+1)'}{2(x^2+1)^2}\\
& u' = 1 \quad v = -\frac{1}{2} \frac{1}{x^2+1}
\end{split}
\end{equation} \Biggr\rvert$
$= \arctan x - (-x\frac{1}{2}\frac{1}{x^2+1} + \frac{1}{2} \int \frac{dx}{x^2+1})$
$= \arctan x + \frac{x}{2(x^2+1)} - \frac{1}{2}\arctan x + C = \frac{1}{2}\arctan x + \frac{x}{2(x^2+1)} + C$
Thank you.

I used:
$$p=x^2+1\to dx =\frac{1}{2\sqrt{p-1}}$$
which leads to:
$$\int{\frac{dx}{(x^2+1)^3}=\frac12 \int{\frac{dp}{p\sqrt{p-1}}}}$$
Then the substitution $$q=\sqrt{p-1}\to dp=2q\space dq$$
Leads to $$\frac12\int\frac{dp}{p\sqrt{p-1}}=\frac12\int\frac{2q}{q(q^2+1)}dq=\int\frac{dq}{q^2+1}$$
Which is resolved simply.
