How does one prove $\int_0^\infty \frac{\log(x)}{1 + e^{ax}} \, dx = -\frac{\log(2)(2\log(a) + \log(2))}{2a}$ for $a > 0$?

Let $I$ be the integral given by

$$\bbox[5px,border:2px solid #C0A000]{I=\int_0^\infty\frac{\log(x)}{1+e^x}\,dx} \tag 1$$

Expanding the denominator of $(1)$ in a series of $e^{-nx}$ and interchanging the order of summation and integration reveals

$$\begin{align} I&=\int_0^\infty\frac{e^{-x}\log(x)}{1+e^{-x}}\,dx\\\\ &=\int_0^\infty\log(x)\sum_{n=0}^\infty (-1)^ne^{-(n+1)x}\,dx\\\\ &=\sum_{n=0}^\infty (-1)^n \int_0^\infty\log(x)e^{-(n+1)x}\,dx\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}\int_0^\infty e^{-x}(\log(x)-\log(n+1))\,dx \\\\ &=\sum_0^\infty \frac{(-1)^{n+1}(\gamma+\log(n+1))}{n+1}\\\\ &=-\gamma\log(2)+\color{blue}{\sum_{n=1}^\infty \frac{(-1)^{n}\log(n)}{n}}\\\\ &=-\gamma\log(2)+\color{blue}{\eta'(1)}\\\\ &=-\gamma\log(2)+\gamma\log(2)-\frac12\log^2(2)\\\\ &=-\frac12\log^2(2) \end{align}$$

where $\eta'(s)$ is the derivative of the Dirichlet Eta Function with $\eta'(1)=\gamma\log(2)-\frac12\log^2(2)$ (SEE THIS ANSWER).

Therefore, we have the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty\frac{\log(x)}{1+e^x}\,dx=-\frac12\log^2(2)}$$

which agrees with that obtained using Wolfram Alpha!

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NOTE $1$: DIRECT EVALUATION OF THE SUM $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}\log(n)}{n}$

Here, we provide a direct evaluation of the sum $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}\log(n)}{n}$ facilitated by use of the Euler-Maclaurin Summation Formula (EMSF). To that end we proceed.

First, we note that we can write for any $N\ge 1$

$$\begin{align} \sum_{n=1}^{2N}\frac{(-1)^{n}\log(n)}{n}&=2\sum_{n=1}^{N}\frac{\log(2n)}{2n}-\sum_{n=1}^{2N}\frac{\log(n)}{n}\\\\ &=\log(2)\sum_{n=1}^N\frac1n -\sum_{n=N+1}^{2N}\frac{\log(n)}{n} \tag 2 \end{align}$$

Using the EMSF, we expand the second sum on the right-hand side of $(2)$ to obtain

$$\begin{align} \sum_{n=N+1}^{2N}\frac{\log(n)}{n}&=\int_N^{2N}\frac{\log(x)}{x}\,dx+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12\log^2(2N)-\frac12\log^2(N)+O\left(\frac{\log(N)}{N}\right)\\\\ &=\log(2)\log(N)+\frac12\log^2(2)+O\left(\frac{\log(N)}{N}\right) \tag 3 \end{align}$$

Substituting $(3)$ in $(2)$ reveals

$$\begin{align} \sum_{n=1}^{\infty}\frac{(-1)^{n}\log(n)}{n}&=\lim_{N\to \infty}\sum_{n=1}^{2N}\frac{(-1)^{n}\log(n)}{n}\\\\ &=\lim_{N\to \infty}\left(\log(2)\sum_{n=1}^n\frac1n -\log(2)\log(N)-\frac12\log^2(2)+O\left(\frac{\log(N)}{N}\right)\right)\\\\ &=\log(2)\lim_{N\to \infty}\left(\sum_{n=1}^N \frac1n -\log(N)\right)-\frac12\log^2(2)\\\\ &=\gamma \log(2)-\frac12\log^2(2) \end{align}$$

Therefore, we find that

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^{\infty}\frac{(-1)^{n}\log(n)}{n}=\gamma \log(2)-\frac12\log^2(2)}$$