Integral $\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}$ Hi I am trying to show$$
I:=\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}.
$$
Thank you.
What a desirable thing to want to prove!  It is a work of art this one.  I wish to prove this in as many ways as we can find. 
Note I tried writing
$$
I=\int_0^\infty \log(1+x^2)\coth \frac{\pi x}{2} \sinh^{-2} \frac{\pi x}{2}\mathrm dx
$$
but this didn't help me much.  We can also try introducing a parameter as follows
$$
I(\alpha)=\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\alpha \pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx,
$$
But this is where I got stuck.  How can we calculate I?  Thanks.

I will solve the general form
\begin{align}\int^\infty_0\frac{x}{x^2+1}\frac{dx}{\sinh(ax)}&=\int^\infty_0\frac{x}{x^2+a^2}\frac{dx}{\sinh(x)}
\\&=\int^\infty_0 \int^\infty_0e^{-at} \frac{\sin(xt)}{\sinh(x)}\,dt \, dx\\&=\int^\infty_0 e^{-at}\int^\infty_0 \frac{\sin(xt)}{\sinh(x)} \, dx\,dt
 \\&=\frac{\pi}{2} \int^\infty_0 e^{-at}\tanh\left(\frac{\pi}{2} t\right)\,dt\\&=\int^\infty_0 e^{-zx}\tanh(x)\,dt \,\,\,\,; z=\frac{2}{\pi}a\\&=\int^\infty_0\frac{e^{-zx}(1-e^{-2x})}{e^{-2x}+1}\,dx\end{align}
By splitting the integral we have
\begin{align} \int^\infty_0\frac{e^{-zx}}{e^{-2x}+1}\,dx &= \sum_{n\geq 0}\int^\infty_0e^{-x(2n+z)}\,dx\\&=\sum_{n\geq 0}\frac{(-1)^n}{2n+z}\\&=\frac{1}{4}\left (\psi \left(\frac{1}{2}+\frac{z}{4}\right)-\psi \left(\frac{z}{4} \right) \right)\end{align}
\begin{align}-\int^\infty_0\frac{-e^{-x(z+2)}}{e^{-2x}+1}\,dx&=-\sum_{n\geq0}\frac{(-1)^n}{z+2+2n}\\&=-\frac{1}{4}\left(-\psi \left(\frac{1}{2}+\frac{z}{4}\right)+\psi\left(1+\frac{z}{4} \right) \right)\end{align}
Hence we have
\begin{align}\int^\infty_0\frac{x}{x^2+1}\frac{dx}{\sinh(ax)}&=\frac{1}{4}\left (2\psi \left(\frac{1}{2}+\frac{z}{4}\right)-\psi \left(1+\frac{z}{4}\right)-\psi \left(\frac{z}{4} \right) \right)\\&=\frac{1}{2}\psi \left(\frac{1}{2}+\frac{z}{4}\right)-\frac{1}{2}\psi \left(\frac{z}{4} \right)-z \\&=\frac{1}{2}\psi \left(\frac{1}{2}+\frac{a}{2\pi}\right)-\frac{1}{2}\psi \left(\frac{a}{2\pi} \right)-\frac{2}{\pi}a \end{align}
Let $a=\pi/2$
\begin{align}\int^\infty_0\frac{x}{x^2+1}\frac{dx}{\sinh(\frac{\pi}{2}x)}&=\frac{1}{2}\psi \left(\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{2}\psi \left(\frac{1}{4} \right)-1\\&=\frac{\pi}{2} \cot(\pi/4)-1\\&=\frac{\pi}{2}-1\end{align}
