Non-Elementary Integration · Superchallenge problem

Using $\dfrac1{\sinh x}=\dfrac{2e^{-x}}{1-e^{-2x}}=2\displaystyle\sum_{n=0}^{\infty}e^{-(2n+1)x}$ and $\displaystyle\sum_{n=0}^{\infty}\dfrac1{(2n+1)^2}=\dfrac{\pi^2}8$, evaluate $\displaystyle\int_0^{\infty}\dfrac{x}{\sinh x}\,dx$.