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).
 A: For any $\alpha>0$, let:
$$ I(\alpha)= \int_{0}^{+\infty}\frac{dx}{\alpha^2+x^2} = \frac{\pi}{2\alpha}.$$
By differentiating both sides with respect to $\alpha$ we get:
$$ \int_{0}^{+\infty}\frac{2\alpha}{(\alpha^2+x^2)^2}\,dx = \frac{\pi}{2\alpha^2} $$
and by evaluating at $\alpha=1$:
$$ \int_{0}^{+\infty}\frac{dx}{(x^2+1)^2} = \color{red}{\frac{\pi}{4}}.$$
A: Use $u=tan x$
$$(1+x^2)^2=(1+tan^2u)^2=(sec^2u)^2=sec^4u$$
$$dx=sec^2u du$$ 
The integral becomes $$\int{\frac{1}{sec^2u}du=\int{cos^2udu}}$$
A: Use $x = \tan t$. Then $(1 + x^2)^2 = \sec^4 t$, and $\dfrac{dx}{dt} = \sec^2 t$, so the integral becomes
\begin{align*}
\int_0^{\pi/2} \cos^2 t \,\mathrm dt = \dfrac 1 2 \times \dfrac {\pi} 2 = \dfrac{\pi}{4}.
\end{align*}
A: Add and subtract $x^2$ from the numerator, you get $\arctan$ on one hand and an $\int \frac{x^2}{(1 + x^2)^2}dx$, which you can do by parts ($u = x$ and $dv = \frac{x}{(1+x^2)^2}dx$)
A: $$\int_0^\infty\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=\lim_{n\to\infty}\int_0^n\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=$$

Substitute $x=\tan(u)$ and $\text{d}x=\sec^2(u)\space\text{d}u$.
So $\left(1+x^2\right)^2=\left(1+\tan^2(u)\right)^2=\sec^4(u)$ and $u=\arctan(x)$.
This gives a new lower bound $u=\arctan(0)=0$ and upper bound $u=\arctan(n)$:

$$\lim_{n\to\infty}\int_0^{\arctan(n)}\cos^2(u)\space\text{d}u=$$

Use:
$$\cos^2(x)=\frac{1+\cos(2u)}{2}$$

$$\frac{1}{2}\lim_{n\to\infty}\left[\int_0^{\arctan(n)}1\space\text{d}u+\int_0^{\arctan(n)}\cos(2u)\space\text{d}u\right]=$$

Substitute $s=2u$ and $\text{d}s=2\space\text{d}u$.
This gives a new lower bound $s=2\cdot0=0$ and upper bound $s=2\arctan(n)$:

$$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\int_0^{2\arctan(n)}\cos(s)\space\text{d}s\right]=$$
$$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\left[\sin(s)\right]_0^{2\arctan(n)}\right]=$$
$$\frac{1}{2}\lim_{n\to\infty}\left[\left(\arctan(n)-0\right)+\frac{\sin(2\arctan(n))-\sin(0)}{2}\right]=$$
$$\frac{1}{2}\lim_{n\to\infty}\left[\arctan(n)+\frac{\sin(2\arctan(n))}{2}\right]=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}$$
A: let $I_n=\int_{0}^{\infty}\frac{1}{(1+x^2)^n}dx$ then $$I_{n+1}=\frac{2n-1}{2n}I_n$$
we know $I_1=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}$, so 
$$I_2=\frac{2(1)-1}{2(1)}I_1=\frac{\pi}{4}$$
A: Here's a pretty solution. Notice that $$ I=\int_{0}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{\partial\Omega}\frac{1}{(1-iz)^2(1+iz)^2}dz$$ where $\Omega$ is the top part of the complex plane. 
Since $f(z)$ decays faster than $\frac{1}{z^2}$,
 $$I=\pi i \lim_{z \to i}\frac{d}{dz}\frac{(z-i)^2}{(1-iz)^2(1+iz)^2}=\frac{\pi}{4}$$.
A: $$
\begin{aligned}
\int \frac{d x}{\left(1+x^2\right)^2} &=-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^2}\right) \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(-\frac{1}{x^2}\right) \frac{1}{1+x^2} d x \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^2}-\frac{1}{x^2}\right) d x \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \tan ^{-1} x+\frac{1}{2 x}+C \\
&=\frac{1}{2}\left(\frac{x}{1+x^2}+\tan ^{-1} x\right)+C
\end{aligned}
$$
For definite integral, we have $$
\begin{aligned}
\int_0^{\infty} \frac{d x}{\left(1+x^2\right)^2} &=\frac{1}{2}\left[\frac{x}{1+x^2}+\tan ^{-1} x\right]_0^{\infty} =\frac{\pi}{4}
\end{aligned}
$$
