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 $x=\pm i$.

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. (>‿◠)✌
