<s>[INST] A closed form for integral $\int_0^\infty \frac{\ln(1-e^{-\pi x})}{1+x^2} dx$ $$\int_0^\infty \frac{\ln(1-e^{-\pi x})}{1+x^2} dx$$
let $u=e^{-\pi x}$, the integral goes to:
$$\pi \int_0^1 \frac{1}{u}\cdot\frac{\ln(1-u)}{\pi^2+\ln^2(u)}du$$
This looks a little like Gregory's coefficient, but not the same, because the integral is from $0$ to $1$, instead of from $0$ to $\infty$. How to do next?
 [/INST] Similar to the evaluation of $\int_{0}^{\infty} \operatorname{Li}_{2}(e^{-\pi x}) \arctan(x) \, \mathrm dx$ here, we have $$ \begin{align} \int_{0}^{\infty} \frac{\ln(1-e^{-\pi x})}{1+x^{2}} \, \mathrm dx &= -\int_{0}^{\infty} \frac{1}{1+x^{2}} \sum_{n=1}^{\infty} \frac{e^{- \pi n x}}{n} \, \mathrm dx\\ &=   -\sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} \frac{e^{- \pi n x}}{1+x^{2}} \, \mathrm dx \\ &= - \sum_{n=1}^{\infty} \frac{1}{n} \left( \left(\frac{\pi}{2} - \operatorname{Si}(\pi n) \right) \cos(\pi n) + \operatorname{Ci}(\pi n ) \sin(\pi n ) \right) \\ &= \frac{\pi}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} \\ &=  \frac{\pi}{2} \ln(2) + \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n}, \end{align}$$
where $\operatorname{Si}(z)$ is the sine integral and $\operatorname{Ci}(z)$ is the cosine integral.
(Justification for interchanging the order of summation and integration comes from Tonelli's theorem.)
To show that $$\sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} = - \frac{\pi}{2}, $$ we can integrate the meromorphic function $$f(z) = \frac{\pi \csc(\pi z) \operatorname{Si}(\pi z)}{z}$$ around a rectangular contour with vertices at $z= \pm  \left(N+ \frac{1}{2} \right) \pm i\left(N+ \frac{1}{2} \right),$ where $N$ is a positive integer.
As $|z| \to \infty$ in $ -\pi < \arg(z) < \pi$ , we have $$f(z) \sim \frac{\pi \csc(\pi z) \left(\frac{\pi}{2}- \frac{\cos(\pi z)}{\pi z}   \right) }{z} = \frac{\pi^{2} \csc(\pi z)}{2z} - \frac{\cot(\pi z)}{z^{2}}.$$
(EDIT: As explained here, the asymptotic expansion of $\operatorname{Si}(z)$ for large complex $z$ can derived from the identity $$ \operatorname{Si}(z) = \frac{\pi}{2} - \cos(z) \int_{0}^{\infty} \frac{e^{-zt}}{1+t^{2}} \, \mathrm dt - \sin(z) \int_{0}^{\infty} \frac{te^{-zt}}{1+t^{2}} \, \mathrm dt \, , \quad \Re(z)  >0$$ using Watson's lemma.)
Therefore, for reasons explained in the answers to this question, the integral vanishes on all sides of the rectangular contour as $N \to \infty$.
So we have $$ \begin{align} 0 &= 2 \pi i \sum_{n=-\infty}^{\infty}\operatorname{Res}[f(z), n]  \\ &= 2 \pi i \left( \sum_{n=-\infty}^{-1} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} + \pi + \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} \right)  \\ &= 2 \pi i \left(2 \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} + \pi \right), \end{align}$$ and the result follows.

Alternatively, $$ \begin{align} \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si} (\pi n)}{n} &= \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \int_{0}^{\pi n} \frac{\sin t}{t} \, \mathrm dt  \\ &=  \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \int_{0}^{\pi} \frac{\sin (n u)}{u} \, \mathrm du  \\ &= \int_{0}^{\pi} \frac{1}{u} \sum_{n=1}^{\infty} (-1)^{n} \frac{\sin (n u)}{n} \, \mathrm du  \\ &= - \int_{0}^{\pi} \frac{1}{u} \, \arctan \left(\tan\frac{ u}{2} \right) \, \mathrm du \\ &= - \int_{0}^{\pi} \frac{1}{u} \frac{u}{2} \, \mathrm du\\ &= - \frac{\pi}{2}. \end{align}$$
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