How to compute $\int \frac{1}{(x^2+1)^2}dx$? Suppose we know $\int \frac{1-x^2}{(x^2+1)^2}=\frac{x}{x^2+1}+C$
How to compute $\int \frac{1}{(x^2+1)^2}dx$?
I tried writing it as $\frac{1+x^2-x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}+\frac{x^2}{(x^2+1)^2}$. But then how do you deal with $\frac{x^2}{(x^2+1)^2}$?
Some ideas?

So your problem reduces to
$$\int\frac{x^2dx}{(x^2+1)^2}$$
Integrate by parts: let $u=x$ and $dv=\frac{x}{(x^2+1)^2}dx$. Then $du=dx$ and $v=-\frac{1}{2}(x^2+1)^{-1}$ so we have
$$-\frac{1}{2}x(x^2+1)^{-1}+\frac{1}{2}\int\frac{dx}{x^2+1}$$
and that last integral may jump out at you as a particular trig derivative
