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?

To find $$\int\frac{x^2dx}{(x^2+1)^2}$$ you may want consider the trigonometric subsitution $x=\tan t$ and $dx=\sec^2t dt$
This gives:
$$\int\frac{\tan^2t \sec^2t dt}{(\tan^2t+1)^2}=\int\frac{\tan^2t \sec^2t dt}{\sec^4t}=\int\frac{\tan^2t  dt}{\sec^2t}=\int \frac{\sin^2t\cos^2t}{\cos^2t}dt=\int \sin^2tdt$$
The last integral is easily evaluated with the reduction formula for $\sin^2t= \dfrac {1 - \cos 2 t} 2$.
