$\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? How to calculate integral $\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? I got this integral by using Abel-Plana formula on series $\sum_{n=0}^\infty \frac{1}{(n+1)^2}$. This integral can be splitted into two integrals with bounds from 0 to 1 and from 1 to infinity and the both integrals converge, so does the sum. I checked with WolframAlpha and the value of the integral is $\frac{-9 + \pi^2}{24}$, but I don't know how to compute it. Also, I tried to write $\frac{2xdx}{(1+x^2)^2}=d\frac{1}{x^2+1}$ and then tried to use partial integration, but didn't succeded.
Any help is welcome. Thanks in advance.

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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x \over \pars{\expo{2\pi x} - 1}\pars{x^{2} + 1}^{2}}
\dd x}
\\[5mm] = &\
{1 \over 4}\bracks{-2\int_{0}^{\infty}{\Im\pars{\bracks{1 + \ic x}^{-2}} \over \expo{2\pi x} - 1}\dd x}
\end{align}
The brackets-$\ds{\bracks{}}$ enclosed expression can be evaluated with the Abel-Plana Formula. Namely,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x \over \pars{\expo{2\pi x} - 1}\pars{x^{2} + 1}^{2}}
\dd x}
\\[5mm] = &\
{1 \over 4}\bracks{\sum_{n = 0}^{\infty}{1 \over \pars{1 + n}^{2}} -
\int_{0}^{\infty}{\dd n \over \pars{1 + n}^{2}} -
\left.{1 \over 2}{1 \over \pars{1 + n}^{2}}\right\vert_{n\ =\ 0}}
\\[5mm] = &
{1 \over 4}\pars{{\pi^{2} \over 6} - 1 - {1 \over 2}} = \bbox[5px,#ffd]{{\pi^{2} \over 24} - {3 \over 8}} \approx 0.0362
\end{align}
