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. =)
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$\ds{\int_{-\infty}^{\infty}{\expo{2x} -\expo{x}
     \over x\pars{1 + \expo{2x}}\pars{1 + \expo{x}}}\,\dd x=\ln\pars{2}:
     \ {\large ?}}$

\begin{align}&\color{#c00000}{%
\int_{-\infty}^{\infty}{\expo{2x} -\expo{x}
\over x\pars{1 + \expo{2x}}\pars{1 + \expo{x}}}\,\dd x}
=2\lim_{\Lambda \to \infty}\int_{0}^{\Lambda}{1 \over x}
\pars{{1 \over 1 + \expo{x}} - {1 \over 1 + \expo{2x}}}\,\dd x
\\[3mm]&=2\lim_{\epsilon \to 0^{+} \atop \Lambda \to \infty}\bracks{%
\int_{\epsilon}^{\Lambda}{1 \over 1 + \expo{x}}\,{\dd x \over x}
-\int_{\epsilon}^{\Lambda}{1 \over 1 + \expo{2x}}\,{\dd x \over x}}
\\[3mm]&=2\lim_{\epsilon \to 0^{+} \atop \Lambda \to \infty}\bracks{%
\int_{\epsilon}^{\Lambda}{1 \over 1 + \expo{x}}\,{\dd x \over x}
-\int_{2\epsilon}^{2\Lambda}{1 \over 1 + \expo{x}}\,{\dd x \over x}}
\\[3mm]&=2\lim_{\epsilon \to 0^{+} \atop \Lambda \to \infty}\bracks{%
\int_{\epsilon}^{2\epsilon}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}
+\int_{2\epsilon}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}
-\int_{\Lambda}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}
-\int_{2\epsilon}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}}
\\[3mm]&=2\lim_{\epsilon \to 0^{+} \atop \Lambda \to \infty}\bracks{%
\half\int_{\epsilon}^{2\epsilon}{\dd x \over x}
+\int_{\epsilon}^{2\epsilon}\pars{{1 \over 1 + \expo{x}} - \half}\,{\dd x \over x}
-\int_{\Lambda}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}}
\\[3mm]&=\ln\pars{2}
-\half\
\overbrace{\lim_{\epsilon \to 0^{+}}
\int_{\epsilon/2}^{\epsilon}{\tanh\pars{x} \over x}\,\dd x}^{\ds{=\ 0}}\
-\
\overbrace{\lim_{\Lambda \to \infty}
\int_{\Lambda}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}}
^{\ds{=\ 0}}\,,
\qquad\pars{~\large\tt\mbox{See below.}~}
\end{align}

$$\color{#00f}{\large%
\int_{-\infty}^{\infty}{\expo{2x} -\expo{x}
     \over x\pars{1 + \expo{2x}}\pars{1 + \expo{x}}}\,\dd x=\ln\pars{2}}
$$

$$
{\tanh\pars{x} \over x} = \sech^{2}\pars{\xi} \leq 1\,,\qquad
0 < \xi < x
$$
  $$
\verts{\int_{\epsilon/2}^{\epsilon}{\tanh\pars{x} \over x}\,\dd x}
<\half\,\epsilon\to 0\quad\mbox{when}\quad\epsilon\to 0^{+}
$$

$$
\verts{\int_{\Lambda}^{2\Lambda}{1 \over 1 + \expo{x}}\,{\dd x \over x}}
<{1 \over \Lambda}\int_{\Lambda}^{2\Lambda}{\expo{-x} \over 1 + \expo{-x}}\,\dd x
={1 \over \Lambda}\ln\pars{1 + \expo{-\Lambda} \over 1 + \expo{-2\Lambda}}
\to 0\quad\mbox{when}\quad\Lambda\to\infty
$$
A: I reckon you are on the right track with Frullani's Integral, although the series expansion of the exponential function in the latter part of your derivation complicates matters unnecessarily. 
Using a partial fraction decomposition and symmetry,
$$\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x =\int_{-\infty}^{\infty} \frac{1}{x}\left[\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right]\mathrm{d}x\\=2\int_{0}^{\infty} \frac{1}{x}\left[\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right]\mathrm{d}x$$
Setting $f(x)=\frac{1}{1+e^x}$, we are in a position to use Frullani's integral as follows
$$I=2\int_0^{\infty}\frac{f(x)-f(2x)}{x}\mathrm{d}x=2[f(0)-f(\infty)]\log\left(\frac{2}{1}\right)=2\left(\frac{1}{2}\right)\log 2=\log 2$$
A: $$\begin{align}\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x &=\int_{-\infty}^{\infty} \frac{1}{x}\left[\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right]\mathrm{d}x\\&=2\int_{0}^{\infty} \frac{1}{x}\left[\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right]\mathrm{d}x\\&=2\lim_{s\to0^+}\int_0^\infty\frac{x^{s-1}\ }{e^x+1}\mathrm dx-\int_0^\infty\frac{x^{s-1}}{e^{2x}+1}\ \mathrm dx\end{align}$$
By letting $x\mapsto\frac12x$ in the last integral, we get
$$\int_0^\infty\frac{x^{s-1}}{e^{2x}+1}\ \mathrm dx=2^{-s}\int_0^\infty\frac{x^{s-1}}{e^x+1}\ \mathrm dx$$
Likewise, using the Dirichlet eta function, we have
$$\int_0^\infty\frac{x^{s-1}\ }{e^x+1}\mathrm dx=\Gamma(s)\eta(s)$$
and so,
$$\begin{align}\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x&=2\lim_{s\to0^+}(1-2^{-s})\Gamma(s)\eta(s)\\&=2\lim_{s\to0^+}\color{red}{\frac{1-2^{-s}}s}\color{blue}{s\Gamma(s)}\eta(s)\\&=2\color{red}{\ln(2)}\color{blue}{(1)}\eta(0)\\&=\ln(2)\end{align}$$
A: 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
$$
\varphi(x)=\frac1{1+\mathrm e^x}.
$$
