How can we show that $\ln{2}=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}\left({12\over e^{n\pi}-1}+{4\over e^{n\pi}+1}\right)$ $$\ln{2}=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}\left({12\over e^{n\pi}-1}+{4\over e^{n\pi}+1}\right)\tag1$$
Any hints?

This is a result of theory of theta functions. Let $0 < q < 1$ and $$a(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})}$$ Next we can see that \begin{align}
b(q) &= \sum_{n = 1}^{\infty}(-1)^{n - 1}\cdot\frac{q^{n}}{n(1 - q^{n})}\notag\\
&= \sum_{n \text{ odd}}\frac{q^{n}}{n(1 - q^{n})} - \sum_{n \text{ even}}\frac{q^{n}}{n(1 - q^{n})}\notag\\
&= \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})} - 2\sum_{n \text{ even}}\frac{q^{n}}{n(1 - q^{n})}\notag\\
&= a(q) - a(q^{2})\notag
\end{align}
Similarly we can prove that $$c(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 + q^{n})} = a(q) - 2a(q^{2})$$ and $$d(q) = \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{q^{n}}{n(1 + q^{n})} = c(q) - c(q^{2})$$ i.e $$d(q) = \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{q^{n}}{n(1 + q^{n})} = a(q) - 3a(q^{2}) + 2a(q^{4})$$ The sum in question is $$S = 12b(q) + 4d(q) = 16a(q) - 24a(q^{2}) + 8a(q^{4})\tag{1}$$ where $q =  e^{-\pi}$. We can see from this post that
\begin{align}
a(q) &= -\frac{\log kk'^{4}}{12} - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{24K}\tag{2}\\
a(q^{2}) &= -\frac{\log kk'}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\tag{3}\\
a(q^{4}) &= -\frac{\log k^{4}k'}{12} + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\tag{4}
\end{align}
Since $q = e^{-\pi}$ it follows from theory of theta functions and elliptic integrals that $K' = K$ and $k = k' = 1/\sqrt{2}$. Thus $$a(q) = -\frac{\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{24}$$ and $$a(q^{2}) = -\frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{12}$$ and $$a(q^{4}) = \frac{3\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{6}$$ and then from $(1)$ we get $S = \log 2$.
