There's a way avoiding special functions and transforms, similar to the device for calculating Frullani integrals: Let's calculate $\displaystyle\int^\infty_\epsilon\frac{\log x}{1+e^x}\,dx$ up to terms $o(1)$ for $\epsilon\to0+$. With the identity $$\frac1{e^x+1}=\frac1{e^x-1}-\frac2{e^{2x}-1},$$ we have
\begin{align}\int^\infty_\epsilon\frac{\log x}{1+e^x}\,dx&=\int^\infty_\epsilon\frac{\log x}{e^x-1}\,dx-\int^\infty_\epsilon\frac{2\log x}{e^{2x}-1}\,dx\\&=\int^\infty_\epsilon\frac{\log x}{e^x-1}\,dx-\int^\infty_{2\epsilon}\frac{\log x-\log2}{e^x-1}\,dx
\\&=\int^{2\epsilon}_\epsilon\frac{\log x}{e^x-1}\,dx+\int^\infty_{2\epsilon}\frac{\log2}{e^x-1}\,dx
\end{align}
Using $\displaystyle\frac1{e^x-1}=\frac1x+O(1)$ and $\displaystyle\int^{2\epsilon}_\epsilon|\log x|\,dx=o(1)$, we see
$$\int^{2\epsilon}_\epsilon\frac{\log x}{e^x-1}\,dx=\int^{2\epsilon}_\epsilon\frac{\log x}{x}\,dx+o(1)=\frac12\log^22+\log2\log\epsilon+o(1)$$
and $$\int^\infty_{2\epsilon}\frac{\log2}{e^x-1}\,dx=\log2\log\frac1{1-e^{-2\epsilon}}=-\log^22-\log2\log\epsilon+o(1),$$ so
$$\int^\infty_\epsilon\frac{\log x}{1+e^x}\,dx=-\frac12\log^22+o(1),$$ and our integral is $$\int^\infty_0\frac{\log x}{1+e^x}\,dx=-\frac12\log^22.$$