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 $x=\pm i$.
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 $a=1$.
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,
$$\begin{align}\int_0^{\infty} dx \: \frac{\sin^2{x}}{x^2} \frac1{1+x^2} &= \frac{\pi}{2} \int_0^2 dk \, \left (1 - \frac{k}{2} \right ) e^{-k}\\ &= \frac{\pi}{2} \left (1-e^{-2}- \frac12 (1-3 e^{-2}) \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align}$$
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. (>‿◠)✌
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 $$ I=\frac{\pi}{4}\int_0^1\int_0^1(e^{-(a+b)}+e^{-|a-b|})dadb=\frac{(1+e^2)\pi}{4e^2}. $$
Just a bit late for an interesting integral.
$$I=\int \frac{\sin^2(x)}{x^2(x^2+1)}\,dx=\frac 12 \int \frac{1-\cos(2x)}{x^2(x^2+1)}\,dx$$ $$I=\frac{1}{2} \left(\frac{1}{x}+\tan ^{-1}(x)\right)-\frac 12 \int \frac{\cos(2x)}{x^2(x+i)(x-i)}\,dx$$
Using partial fractions $$\frac{1}{x^2(x+i)(x-i)}=\frac{1}{x^2}-\frac{i}{2 (x+i)}+\frac{i}{2 (x-i)}$$
$$\int \frac{\cos(2x)}{x^2}\,dx=-2\, \text{Si}(2 x)-\frac{\cos (2 x)}{x}$$
$$J_a=\int \frac{\cos(2x)}{x+a}\, dx$$$$x=t-a \quad \implies \quad J_a=\cos(2a)\int \frac{\cos(2t)}{t}\, dt+\sin(2a)\int \frac{\sin(2t)}{t}\, dt$$
Combining all the above gives a bunch of terms. Using the bounds $$\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)}\,dx=\frac{1}{4} \left(3+\frac{1}{e^2}\right) \pi-\frac \pi 2=\frac{\left(1+e^2\right) \pi }{4 e^2}$$
