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.

As Lucian stated in the comments, integrating by parts shows that the integral is equivalent to showing that $$ \int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{\sinh \frac{\pi x}{2}} \ dx = \frac{\pi}{2}-1 .$$
Let $ \displaystyle f(z) = \frac{z}{1+z^{2}} \frac{1}{\sinh \frac{\pi z}{2}} $
and integrate around a rectangle with vertices at $\pm N$ and $\pm N+ i (2N+1)$ where $N$ is some positive integer.
As $N$ goes to infinity through the integers, the integral vanishes on the left and right sides of the rectangle and along the top of the rectangle.
In particular, the absolute value of the integral along the top of the rectangle is bounded by $$\frac{3N+1}{(2N+1)^{2}-1}\int_{-\infty}^{\infty} \frac{1}{\cosh \frac{\pi x}{2}} \ dx = \frac{6N+2}{(2N+1)^{2}-1} \to 0 \ \text{as} \ N \to \infty .$$
Then
$$\int_{-\infty}^{\infty} \frac{x}{1+x^{2}} \frac{1}{\sinh \frac{\pi x}{2}} \ dx = 2 \pi i \left(\text{Res}[f(z),i]+ \sum_{k=1}^{\infty} \text{Res}[f(z),2ki] \right)$$
where
$$ \text{Res}[f(z),i] = \lim_{z \to i} \frac{z}{z+i} \frac{1}{\sinh \frac{\pi x}{2}} =\frac{1}{2i}$$
and
$$ \text{Res}[f(z),2ki] = \lim_{z \to 2ki} \frac{z}{2z \sinh \frac{\pi z}{2}+(1+z^{2}) \frac{\pi}{2} \cosh \frac{\pi z}{2}} = \frac{4i}{\pi} \frac{(-1)^{k} k}{1-4k^{2}} . $$
And notice that
$$ \begin{align} \sum_{k=1}^{\infty} \frac{(-1)^{k} k}{1-4k^{2}} &= -\frac{1}{4} \sum_{k=1}^{\infty} \left( \frac{(-1)^{k}}{2k+1} + \frac{(-1)^{k}}{2k-1} \right) \\ &= -\frac{1}{4} \left(\arctan(1)-1 - \arctan(1) \right) \\ &= \frac{1}{4} . \end{align}$$
Therefore,
$$ \int_{-\infty}^{\infty} \frac{x}{1+x^{2}} \frac{1}{\sinh \frac{\pi x}{2}} \ dx = 2 \pi i \left(\frac{1}{2i} + \frac{i}{\pi} \right) = \pi - 2$$
which implies 
$$  \int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{\sinh \frac{\pi x}{2}} \ dx = \frac{\pi}{2} -1 .$$
