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?

A solution using contour integration. For "small" $r>0$, let $D_r$ be the disk $|z|<1$ with "notches of size $r$" around the point $z=1$ and the negative real axis. Formally, say $$D_r=\{z\in\mathbb{C}:r<|z|<1\land|1-z|>r\land(\Re z>0\lor\Im z>r)\}.$$ Its boundary is $\partial D_r=C_r^0\cup C_r^1\cup C_r^+\cup C_r^-\cup L_r^+\cup L_r^-$, where $C_r^0$ and $C_r^1$ are arcs of radius $r$ around $z=0$ and $z=1$, $C_r^+$ and $C_r^-$ are arcs of $|z|=1$ in the upper/lower half-plane, and finally $L_r^+$ and $L_r^-$ are line segments. Then $\int_{\partial D_r}f(z)\,dz=0$, where $f(z)=\frac{\log(1+z)}{z\log z}$ with the principal values taken. But (using the "half-residue" trick for $C_r^1$)
\begin{align*}
\lim_{r\to 0}&\int_{C_r^0}f(z)\,dz=0,\\
\lim_{r\to 0}&\int_{C_r^1}f(z)\,dz=-i\pi\operatorname*{Res}_{z=1}f(z)=-i\pi\log2,\\
\lim_{r\to 0}&\int_{C_r^+\cup C_r^-}f(z)\,dz=\int_0^\pi\frac{\log(1+e^{it})-\log(1+e^{-it})}{t}dt=i\pi,\\
\lim_{r\to 0}&\int_{L_r^+\cup L_r^-}f(z)\,dz=\int_0^1\frac{\log(1-t)}{t}\left(\frac1{\log t-i\pi}-\frac1{\log t+i\pi}\right)dt=2iI
\end{align*}
where $I=\pi\int_0^1\frac{\log(1-t)}{\pi^2+\log^2 t}\frac{dt}{t}$ is the given integral (as in the OP). Hence $I=\frac\pi2(\log2-1)$.
