<s>[INST] Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$ Is it possible to confirm the value of this integral using the methods of complex analysis or similar? 
$$
\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $\zeta$-function, but I was looking for an explicit calculation. 
 [/INST] This will require some calculations but no special functions. Making the change of variables $x = -x$ on $\mathbb R^-$, we obtain
$$\int_{\mathbb R} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx =
 \underbrace {\int_{\mathbb R^+} \frac {2 x} {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx}_
  {= I} +
 \int_{\mathbb R^+} \frac x {(x^2 + 1)^2} dx, \\
I = 2 \pi i \sum_{k \geq 1}
 \operatorname*{Res}_{x = i k} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} +
 \frac 1 {2 (x^2 + 1)} \bigg\rvert_{x = 0}^\infty = \\
2 \pi i \frac {4 \pi^2 + 3} {96 \pi i} -
 2 \pi i \sum_{k \geq 2} \frac k {2 \pi i (k^2 - 1)^2} -
 \frac 1 2 = \\
\frac {4 \pi^3 - 21} {48} - \frac 1 4 \sum_{k \geq 2}
 \left( \frac 1 {(k - 1)^2} - \frac 1 {(k + 1)^2} \right) = \\
\frac {4 \pi^3 - 21} {48} - \frac 1 4 \left( 1 + \frac 1 4 \right) =
\frac {\pi^2 - 9} {12}.$$
</s>