Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2}  \right) ^2 dx =  \lim_{ A \to \infty }  \int_{0}^{A} \left( \frac{x}{1+x^2}  \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2}  \right) ^2 dx $ but I don't have idea.
 A: As the denominator contains $x^2+1$ we put  $x=\tan y\implies dx=\sec^2ydy$
For $x^2-1$ with $x\ge1$ we need to put $x=\sec y$ 
and for $1-x^2$ with $x\le1$ we need to put $x=\sin y$ 
$$\implies \int\frac{x^2dx}{(x^2+1)^2}=\int\frac{\tan^2y\sec^2ydy}{\sec^4y}=\int\sin^2ydy=\frac12\int(1-\cos2y)dy=\frac{y}2-\frac{\sin2y}4+C$$
If it were indefinite integral, we had to replace back $y$ with $x$
$y=\arctan x$ and $\sin2y=\frac{2\tan y}{1+\tan^2y}=\frac{2x}{1+x^2}$
For definite integral, $x=0,y=\arctan 0=0$ and $x=\infty,y=\arctan \infty=\frac\pi2$
So, the required definite integral will be $$2\left[\frac{y}2-\frac{\sin2y}4+C\right]_0^{\frac\pi2}=2\cdot\frac\pi4=\frac\pi2$$
A: $$\frac{\pi}{t}=\int_{-\infty}^{\infty} \frac{dx}{t^2x^2+1}\Rightarrow\frac{\pi}{t^2}=\int_{-\infty}^{\infty}\frac{2tx^2}{(t^2x^2+1)^2}\,dx\Rightarrow \frac{\pi}{2}=\int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)^2}\,dx$$
A: $$F(a)=\int-\frac{1}{1+ax^2}dx=-\frac{\arctan(\sqrt{a}x)}{\sqrt a}+c$$
$$\int_{-\infty}^{+\infty}-\frac{1}{1+ax^2}dx=-\frac{\pi}{\sqrt a}$$
so we take the derivaitve of $-\frac{\pi}{\sqrt a}$ with respect to $"a"$ then put a=1 so 
$$\int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=\frac{\pi}{2}$$
A: HINT:
$$I=\int\frac{x^2dx}{(x^2+1)^2}=\int x\cdot \frac x{(x^2+1)^2}dx$$
Integrating by parts, 
$$I=x\int \frac x{(x^2+1)^2}dx-\int\left(\frac{d(x)}{dx}\cdot \frac x{(x^2+1)^2}dx\right)dx$$
$$\text{As }\int \frac x{(x^2+1)^2}dx=\frac12\int\frac{d(x^2+1)}{(x^2+1)^2}=-\frac1{2(x^2+1)}$$
$$I=-x\cdot\frac1{2(x^2+1)}+\int \frac1{2(x^2+1)}dx=-\frac x{2(x^2+1)}+\frac{\arctan x}2+C $$
Can you take it from here?
A: This integral may be evaluated using the residue theorem.  The integrand has poles at $x=\pm i$; if we close with a contour in the upper half plane, then we need only worry about the pole at $x=i$.  The integral is therefore
$$i 2 \pi \left [\frac{d}{dx} \frac{x^2}{(x+i)^2} \right ]_{x=i} = i 2 \pi\left [\frac{i 2 x}{(x+i)^3} \right ]_{x=i} = i 2 \pi \frac{-i}{4} = \frac{\pi}{2}$$
A: I know I'm a bit late for the party, but here is another solution:
$$\begin{align}
2\int_{0}^\infty \frac{x^2}{(1+x^2)^2}\mathrm dx &=
\frac22\int_{-\infty}^\infty \frac{y^{\frac{2-1}{2}}}{(1+y)^2}\mathrm dx \\
&= B\bigl(\tfrac12 + 1, 2-\tfrac12 -1\bigr) = B\bigl(\tfrac32,\tfrac12\bigr) \\
&= \frac{\Gamma\bigl(\tfrac32\bigr)\Gamma\bigl(\tfrac12\bigr)}{\Gamma(2)} \\
&= \frac{\sqrt\pi \cdot \frac12 \sqrt\pi}{1} = \frac\pi 2
\end{align}
$$
The first step is the substitution $x^2 \mapsto y$, the second and third step are the use of well-known (and very easy to proof) representations of the Beta function.
A: We can use Parseval's theorem!
If $F$ is the Fourier transform of $f$, and $G$ is the Fourier transform of $g$, then
$$
\int_{-\infty}^\infty\overline{f(t)}g(t)\,dt=
\frac{1}{2\pi}\int_{-\infty}^\infty\overline{F(x)}G(x)\,dx.
$$
We have the Fourier transform pairs


*

*$\displaystyle e^{-|t|}\longmapsto\frac{2}{1+\omega^2}$

*$\displaystyle \frac{d}{dt}f(t) \longmapsto i\omega F(\omega)$


It is clear from those that
$$
f(t)=\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\longmapsto
\frac{ix}{1+x^2}=F(x),
$$
and conveniently $\overline{F(x)}F(x)=\frac{x^2}{(1+x^2)^2}$ is the integrand that we want.
Using this and Parseval's theorem, we see that
$$
\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx=
2\pi\int_{-\infty}^\infty\left(\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\right)^2dt,
$$
and since the integrand is even, one can write
$$
\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx=
\pi\int_0^\infty\left(\frac{d}{dt}\left(e^{-t}\right)\right)^2dt=
\pi\int_0^\infty e^{-2t}dt=\frac{\pi}{2}.
$$
