integration - Prove that $int_{-infty}^{infty} frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}mathrm{d}x=log 2$

This integral popped up recently

$$\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x = \log 2$$

A solution using both real and complex analysis is welcome. I tried rewriting it using symmetry, and then the series expansion of $1/(1+e^{nx})$, however this did not quite make it.

$$\begin{align*} I & = 2\int_{0}^{\infty} \frac{e^{2x}-e^x}{x e^{3x}(1+e^{-2x})(1+e^{-x})}\mathrm{d}x \\ & = 2 \int_{0}^{\infty}\frac{e^{-2x} -e^{-x}}{x} \left(\sum_{n=0}^\infty (-1)^n e^{-n x}\right)\left(\sum_{m=0}^\infty (-1)^me^{-2mx}\right) \mathrm{d}x \end{align*}$$
The first part reminds me of a Frullani integral (it evaluates to $\log 2$). However I am unsure if this is the correct path, any help would be appreciated. =)

Answer

One can start from the identity
$$\frac{\mathrm e^{2x}-\mathrm e^x}{(1+\mathrm e^{2x})(1+\mathrm e^{x})}=\frac1{1+e^{x}}-\frac1{1+\mathrm e^{2x}}=-\int_1^2\frac{\mathrm d}{\mathrm du}\left(\frac1{1+\mathrm e^{ux}}\right)\cdot\mathrm du,$$
that is,

$$\frac{\mathrm e^{2x}-\mathrm e^x}{x(1+\mathrm e^{2x})(1+\mathrm e^{x})}=\int_1^2\frac{\mathrm e^{ux}}{(1+\mathrm e^{ux})^2}\mathrm du.$$
Note that, for every nonzero $u$,
$$\int_{-\infty}^\infty\frac{\mathrm e^{ux}}{(1+\mathrm e^{ux})^2}\mathrm dx=\left.\frac{-1}{u(1+\mathrm e^{ux})}\right|_{-\infty}^\infty=\frac1{|u|},$$
hence Fubini theorem shows that the integral to be computed is
$$\int_{-\infty}^{\infty} \frac{\mathrm e^{2x}-\mathrm e^{x}}{x (1+\mathrm e^{2x})(1+\mathrm e^{x})}\mathrm{d}x = \int_1^2\frac{\mathrm du}u=\log2.$$
More generally, for every positive $a$ and $b$,
$$\int_{-\infty}^{\infty} \frac{\mathrm e^{ax}-\mathrm e^{bx}}{x (1+\mathrm e^{ax})(1+\mathrm e^{bx})}\mathrm{d}x = \log\left(\frac{a}b\right).$$
This is rediscovering the fact that, for every monotonous function $\varphi$ with limits at $\pm\infty$,
$$\int_\mathbb R\frac{\varphi(ax)-\varphi(bx)}x\mathrm dx=(\varphi(+\infty)-\varphi(-\infty))\cdot\log\left(\frac{b}a\right).$$

in the present case for the function
$$\varphi(x)=\frac1{1+\mathrm e^x}.$$