How to integrate $\int \frac{dx}{(1+x^2)^2}$? I need to integrate this to finish an old STEP problem I'm doing, but I'm stuck here, at the very end:
$$\int_0^\infty \frac{dx}{(1+x^2)^2}$$
The result should be $\pi\over 4$ . I don't know how to approach this. *Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).
Also, Wolfram tells me:
$$\int \frac{dx}{(1+x^2)^2} = \frac{1}{2}\left(\frac{x}{x^2+1}+\tan^{-1}x\right)+c$$
but I don't see how one can derive this without knowing the result beforehand.
Please, help me!
Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).
EDIT: If you're interested, the problem in question is: STEP II - problem 4 (year 2014).

$$
\begin{aligned}
\int \frac{d x}{\left(1+x^2\right)^2} &=-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^2}\right) \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(-\frac{1}{x^2}\right) \frac{1}{1+x^2} d x \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^2}-\frac{1}{x^2}\right) d x \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \tan ^{-1} x+\frac{1}{2 x}+C \\
&=\frac{1}{2}\left(\frac{x}{1+x^2}+\tan ^{-1} x\right)+C
\end{aligned}
$$
For definite integral, we have $$
\begin{aligned}
\int_0^{\infty} \frac{d x}{\left(1+x^2\right)^2} &=\frac{1}{2}\left[\frac{x}{1+x^2}+\tan ^{-1} x\right]_0^{\infty} =\frac{\pi}{4}
\end{aligned}
$$
