Differentiation wrt parameter $\int_0^\infty \sin^2(x)\cdot(x^2(x^2+1))^{-1}dx$ Use differentiation with respect to parameter obtaining a differential equation to solve
$$
\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)}dx
$$
No complex variables, only this approach.  Interesting integral and it should have a nice ODE.  I have not found the right way yet.  we have singularities at $x=\pm i$.
 A: Well, as an alternative (like Mr. Ron Gordon did).
\begin{align}
\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\
&=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\
&=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos2x}{1+x^2}dx\\
&=\frac{\pi}{2}-\frac{1}{2}\frac{\pi}{2}+\frac{1}{2}\frac{\pi}{2e^2}\\
&=\frac{\pi}{4}+\frac{\pi}{4e^2}
\end{align}
where I use these links: $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx$ and $\displaystyle\int_0^\infty\frac{\cos2x}{1+x^2}dx$ to help me out.
Unfortunately, this is not differentiation with respect to parameter method (the Feynman way) but I still love this method. (>‿◠)✌
A: How about this method? Denote the integral by $I$. Let
$$ I(a,b)=\int_0^\infty \frac{\sin(ax)\sin(bx)}{x^2(1+x^2)}dx. $$
Then $I(0,b)=I(a,0)=0$, $I(1,1)=I$ and
\begin{eqnarray}
\frac{\partial^2I}{\partial a\partial b}&=&\int_0^\infty\frac{\cos(ax)\cos(bx)}{1+x^2}dx\\
&=&\frac{1}{2}\int_0^\infty\frac{\cos((a+b)x)}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos((a-b)x)}{1+x^2}dx\\
&=&\frac{\pi}{4}(e^{-(a+b)}+e^{-|a-b|})
\end{eqnarray}
where we use
$$ \int_0^\infty\frac{\cos(\alpha x)}{1+x^2}dx=\frac{\pi}{2}e^{-|\alpha|}. $$
Thus
$$ I=\frac{\pi}{4}\int_0^1\int_0^1(e^{-(a+b)}+e^{-|a-b|})dadb=\frac{(1+e^2)\pi}{4e^2}. $$
A: Consider for $a>0$
$$
I(a)=\int_0^\infty\frac{\sin^2(ax)}{x^2(x^2+1)}dx
$$
Differentiate it twice. Since
$$
\int_0^\infty\frac{\cos(kx)}{x^2+1}dx=\frac{\pi}{2e^k}
$$
for $k>0$ we get $I''(a)=\pi e^{-2a}$. Note that $I'(0)=I(0)=0$, so after solving respective IVP we get
$$
I(a)=\frac{\pi}{4}(-1+2a+e^{-2a})
$$
It is remains to substitute $a=1$.
A: Well, as an alternative (and I promise, none of the dreaded complex analysis stuff), we could use Parseval's theorem for Fourier transforms:
For example, the FT of $(\sin{x}/x)^2$ is 
$$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2  \\ 0 & |k| > 2  \end{cases}$$
The FT of $1/(1+x^2)$ is
$$\int_{-\infty}^{\infty} dx \: \frac1{1+x^2} e^{i k x} = \pi \, e^{-|k|}$$
By Parseval's theorem,
$$\begin{align}\int_0^{\infty} dx \: \frac{\sin^2{x}}{x^2} \frac1{1+x^2} &= \frac{\pi}{2} \int_0^2 dk \, \left (1 - \frac{k}{2} \right ) e^{-k}\\ &= \frac{\pi}{2} \left (1-e^{-2}- \frac12 (1-3 e^{-2}) \right  )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align}$$
