Proving $\int_0^\infty\frac{e^x-1}{x(e^x+1)^2}dx=6\ln A-\frac12-\frac16\ln2-\frac12\ln\pi$ I want to prove $$I=\int_0^\infty\frac{e^x-1}{x(e^x+1)^2}dx=6\ln A-\frac12-\frac16\ln2-\frac12\ln\pi$$
where $A$ denotes the Glaisher's constant.
My attempt:
Knowing that $$\int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}dx=\ln b-\ln a$$
So I tried $$I=\int_0^\infty\frac{e^x-1}x\sum_{n=0}^\infty(-1)^{n+1}ne^{-(n+1)x}dx$$
But unfortunately $f_n(x)=(-1)^{n+1}ne^{-(n+1)x}$ does't uniform converge in $[0,+\infty)$ and I can't go further.
(I also noticed the relation between integrals containing $e^x\pm1$ and $\zeta$ function. $\zeta'(-1)=\frac1{12}-\ln A$. So I added "riemann-zeta" in the tags)
 A: Let us consider $$I\left(s\right)=\int_{0}^{\infty}x^{s-1}\frac{e^{x}-1}{\left(e^{x}+1\right)^{2}}dx,\,s>1$$ then $$I\left(s\right)=\int_{0}^{\infty}x^{s-1}\left(e^{x}-1\right)\sum_{n\geq1}\left(-1\right)^{n+1}ne^{-x\left(n+1\right)}dx$$ and in this case we can switch the integral with the series, so $$I\left(s\right)=\sum_{n\geq1}\left(-1\right)^{n+1}n\int_{0}^{\infty}x^{s-1}\left(e^{x}-1\right)e^{-x\left(n+1\right)}dx$$ $$=\Gamma\left(s\right)\sum_{n\geq1}\left(-1\right)^{n+1}n\left(\frac{1}{n^{s}}-\frac{1}{\left(n+1\right)^{s}}\right)$$ and so, rearranging the sum and using the relation $$\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{n^{s}}=\left(1-2^{1-s}\right)\zeta\left(s\right)$$ we obtain $$I\left(s\right)=\Gamma\left(s\right)\left(2^{1-s}\zeta\left(s\right)-\zeta\left(s\right)+2\zeta\left(s-1\right)-2^{3-s}\zeta\left(s-1\right)\right).$$ Now we can note that $I(s)$ is defined also for $s>0$ and, as $s\rightarrow0$, we have a indeterminate form so, by the De L'Hopital's rule, we get \begin{eqnarray*}I\left(0\right) &=& \lim_{s\rightarrow0}\frac{\left(2^{1-s}\zeta\left(s\right)-\zeta\left(s\right)+2\zeta\left(s-1\right)-2^{3-s}\zeta\left(s-1\right)\right)^{\prime}}{\left(1/\Gamma\left(s\right)\right)^{\prime}}\\ &=& \color{red}{6\log\left(A\right)-\frac{1}{2}-\frac{1}{6}\log\left(2\right)-\frac{1}{2}\log\left(\pi\right) }
\end{eqnarray*} as wanted.
