Integration method for $\int_0^\infty\frac{x}{(e^x-1)(x^2+(2\pi)^2)^2}dx=\frac{1}{96} - \frac{3}{32\pi^2}.$ The following definite integral is obtained directly from Hermite's integral representation of the Hurwitz zeta function. But is it possible to obtain the same result via the residue calculus or another technique?
$$\int_{0}^{\infty}
{x
 \over
 \left({\rm e}^{x} - 1\right)\left[x^{2} + \left(2\pi\right)^{2}\right]^{2}}
\,{\rm d}x
={1 \over 96} - {3 \over 32\pi^{2}}.
$$

Binet's second Log Gamma integral formula is 
$$
 \ln \Gamma(z)=\left(z-\frac{1}{2}\right)\ln z-z+\frac{1}{2}\ln(2\pi)+2\int_0^{\infty} \frac{\tan^{-1}(t/z)}{e^{2\pi t}-1} \ \mathrm{d}t 
$$
Differentiating twice w.r.t. $z$ gives 
$$
\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + 4\int_0^{\infty} \frac{zt}{(z^2+t^2)^2(e^{2\pi t}-1)} \ \mathrm{d}t. 
$$
Let $t \mapsto x/2\pi$, then
$$
\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + \int_0^{\infty} \frac{zx}{\pi^2(z^2+x^2/4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x,
$$
$$
\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + 16\pi^2\int_0^{\infty} \frac{zx}{(4\pi^2z^2+x^2)^2(e^{x}-1)} \ \mathrm{d}x. 
$$
Evaluate at $z=1$, 
$$
{\pi^2 \over 6} = \frac{3}{2}+16\pi^2\int_0^{\infty} \frac{x}{(x^2+4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x, 
$$
or, rearranging - 
$$
\int_0^{\infty} \frac{x}{(x^2+4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x = \frac{1}{96} - \frac{3}{32\pi^2}.
$$
