Surely You're Joking, Mr. Feynman! $\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$ 
Prove the following
\begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx=\frac{\pi}{4}+\frac{\pi}{4e^2}\end{equation}

I would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods are welcome). Thank you. (>‿◠)✌

Original question:

And of course, for the sadist with a background in differential
equations, I invite you to try your luck with the last integral of the
group.
\begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx\end{equation}

Source: Integration: The Feynman Way

This integral is readily evaluated using Parseval's theorem for Fourier transforms.  (I am certain that Feynman had this theorem in his tool belt.)  Recall that, for transform pairs $f(x)$ and $F(k)$, and $g(x)$ and $G(k)$, the theorem states that
$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$
In this case, $f(x) = \frac{\sin^2{x}}{x^2}$ and $g(x) = 1/(1+x^2)$.  Then $F(k) = \pi (1-|k|/2) \theta(2-|k|)$ and $G(k) = \pi \, e^{-|k|}$. ($\theta$ is the Heaviside function, $1$ when its argument is positive, $0$ when negative.)  Using the symmetry of the integrand, we may conclude that
$$\begin{align}\int_0^{\infty} dx \frac{\sin^2{x}}{x^2 (1+x^2)} &= \frac{\pi}{2} \int_0^{2} dk \, \left ( 1-\frac{k}{2} \right ) e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) - \frac{\pi}{4} \int_0^{2} dk \, k \, e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) + \frac{\pi}{2 e^2}  - \frac{\pi}{4} \left (1-\frac1{e^2} \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align} $$
