integration - Evaluate $int_0^{infty}frac{log x}{1+e^x},dx$

Evaluate
$$\int_0^{+\infty}\frac{\log x}{1+e^x}\,dx.$$

I have tried using Feynman's Trick (in several ways, but for example by introducing a variable $a$ such that $I(a)=\int_0^{+\infty}\frac{\log ax}{1+e^x}\,dx$), but that doesn't seem to work. Also integration by parts and all kinds of substitutions make things worse (I have no idea how to substitute such that $\log$ and $\exp$ both become simpler.

(Source: Dutch Integration Championship 2013 - Level 5/5)

Answer

By the inverse Laplace transform
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx $$
and by differentiating both sides with respect to $s$
$$ \sum_{n\geq 1}\frac{(-1)^n \log n}{n^s} = -\frac{\Gamma'(s)}{\Gamma(s)^2}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx + \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}\log(x)}{e^x+1}\,dx $$
so by evaluating at $s=1$
$$\int_{0}^{+\infty}\frac{\log x}{e^x+1}\,dx = \sum_{n\geq 1}\frac{(-1)^n\log n}{n}+\underbrace{\Gamma'(1)}_{-\gamma}\underbrace{\int_{0}^{+\infty}\frac{dx}{e^x+1}}_{\log 2} $$
and it just remains to crack the mysterious series $\sum_{n\geq 1}\frac{(-1)^n\log n}{n}$. On the other hand by Frullani's integral, the inverse Laplace transform or Feynman's trick we have $\log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx$, so

$$\sum_{n\geq 1}\frac{(-1)^n\log n}{n}=\int_{0}^{+\infty}\frac{\log(1+e^{-x})-e^{-x}\log 2}{x}\,dx=\gamma\log(2)-\frac{1}{2}\log^2(2)\tag{J}$$
where the last identity follows from the integral representation for the Euler-Mascheroni constant, got by applying the inverse Laplace transform to the series definition $\gamma=\sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]$. Summarizing, we simply have
$$ \int_{0}^{+\infty}\frac{\log(x)}{e^x+1}\,dx = \color{red}{-\frac{1}{2}\log^2(2)}.$$
It is possible to prove the equality between the LHS and the RHS of $(J)$ by summation by parts and Euler sums, too.