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. 

This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I \equiv \int \limits_0^\infty \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)} \, \mathrm{d}x = \int \limits_0^\infty \frac{2 x}{(1+x^2)(\mathrm{e}^{2\pi x}-1)} \, \mathrm{d} x \, . $$
We can then use the geometric series, integration by parts and this Laplace transform to obtain
\begin{align}
I &= \sum \limits_{n=1}^\infty \int \limits_0^\infty \frac{2 x}{(1+x^2)^2} \mathrm{e}^{-2 \pi n x} \, \mathrm{d} x = \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \int \limits_0^\infty \frac{\mathrm{e}^{-2 \pi n x}}{1+x^2} \, \mathrm{d} x \right] \\
&= \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \left(\frac{\pi}{2} - \operatorname{Si}(2 \pi n)\right)\right] \, ,
\end{align}
where $\operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at
$$ I = \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \int \limits_1^\infty \frac{\sin(2 \pi n t)}{t} \, \mathrm{d} t \right] = 2 \sum \limits_{n=1}^\infty \frac{1}{2 \pi n} \int \limits_1^\infty \frac{\sin(2 \pi n t)}{t^3} \, \mathrm{d} t \, .$$
Now the Fourier series
$$ 1 - 2 \{t\} = 4 \sum \limits_{n=1}^\infty \frac{\sin(2 \pi n t)}{2 \pi n} \, , \, t \in \mathbb{R} \, , $$
and the fractional part integral
$$ \int \limits_0^1 x \left\{\frac{1}{x}\right\} \, \mathrm{d} t = \int \limits_1^\infty \frac{\{t\}}{t^3} \, \mathrm{d} t = 1 - \frac{\zeta(2)}{2} = 1 - \frac{\pi^2}{12} $$
are sufficient to derive the final result
$$ I = \frac{1}{2} \int \limits_1^\infty \frac{1 - 2\{t\}}{t^3} \, \mathrm{d} t = \frac{\pi^2}{12} - \frac{3}{4} = \frac{\pi^2 - 9}{12} $$
and thus prove $\pi > 3$ in a rather convoluted manner.
