<s>[INST] One Step in an Integration I was doing the integration
$$
\int\frac{1}{(u^2+a^2)^2}du
$$
and I had a look at the lecturer's steps where I got stuck in the following step:
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
I guess it is integrating this by parts, but I could't see the trick. Could somebody help me with this please?
I had some work after I asked this question here. Here is my approach. 
First note
$$
d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{2a^2u}{(u^2+a^2)^2}du,
$$
then we have
$$
\int\frac{1}{(u^2+a^2)^2}du=\int\left(-\frac{1}{2a^2u}\right)\left(-\frac{2a^2u}{(u^2+a^2)^2}\right)du=\frac{1}{2a^2}\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right).
$$
Then integrate by parts, we have
$$
\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{a^2}{u(u^2+a^2)}-\int\frac{a^2}{u^2(u^2+a^2)}du.
$$
But notice
$$
\frac{a^2}{u^2(u^2+a^2)}=\frac{1}{u^2}-\frac{1}{u^2+a^2},
$$
then we get
$$
\int\frac{a^2}{u^2(u^2+a^2)}du=-\frac{1}{u}-\int\frac{1}{u^2+a^2}du,
$$
then
$$
\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{a^2}{u(u^2+a^2)}+\frac{1}{u}+\int\frac{1}{u^2+a^2}du.
$$
But
$$
\frac{1}{u}-\frac{a^2}{u(u^2+a^2)}=\frac{u}{u^2+a^2},
$$
then
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
Is this approach okay? I am positive with this but not sure it is comprehensive. 
 [/INST] Let $$I(a)= \int \frac{1}{a+x^2} = \frac{1}{\sqrt{a}}\arctan(\frac{x}{\sqrt{a}})$$
Then, 
$$I'(a) = \int \frac{-1}{(a+x^2)^2} = d/da( \frac{1}{\sqrt{a}}\arctan(\frac{x}{\sqrt{a}}))  $$
Find the expression in RHS  and then put $a=u^2$. And you are done! :)
</s>