Integral $I=\int_0^\infty \frac{e^{\alpha x}-e^{\beta x}}{x(e^{\alpha x}+1)(e^{\beta x}+1)}dx, \ \ \alpha>\beta>0. $ $$
I(\alpha,\beta)=\int_0^\infty \frac{e^{\alpha x}-e^{\beta x}}{x(e^{\alpha x}+1)(e^{\beta x}+1)}dx, \ \ \alpha>\beta>0.
$$
I am trying to solve this integral.  This is from the old high school days in Bulgaria, although I cannot find the solutions anymore. Thanks
 A: Split the integrand:
$$\frac{e^{\alpha x}-e^{\beta x}}{(e^{\alpha x}+1)(e^{\beta x}+1)}=
\frac{1}{(e^{\beta x}+1)}-\frac{1}{(e^{\alpha x}+1)}$$
Then recognize this as an integral of another function, evaluated at the limits $\alpha$ and $\beta$. Which function? Well, take the derivative over $\beta$ to find that out:
$$\frac{d}{dy}\left(\frac{1}{x}\frac{1}{(e^{y x}+1)}\right)=-\frac{e^{xy}}{(e^{y x}+1)^2}$$
Your integral becomes
$$I=\int_0^\infty\int_{\beta}^\alpha \frac{e^{xy} dy\,dx}{(e^{xy}+1)^2}$$
Exchange the order of integration:
$$I=\int_{\beta}^\alpha\int_0^\infty \frac{e^{xy} dx\,dy}{(e^{xy}+1)^2}$$
The inner integral
$$\int_0^\infty \frac{e^{xy} dx}{(e^{xy}+1)^2}$$
Becomes
$$\frac{1}{y}\int_0^\infty \frac{e^{u} du}{(e^{u}+1)^2}$$
Use $v=e^u+1$:
$$\frac{1}{y}\int_2^\infty \frac{dv}{v^2}=\frac12\frac{1}{y}$$
Now the outer integral is trivial:
$$I=\frac12\int_{\beta}^\alpha \frac{dy}{y}=\frac12\ln\frac{\alpha}{\beta}$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{{\rm I}\pars{\alpha,\beta}
     =\int_{0}^{\infty}{\expo{\alpha x} -\expo{\beta x} \over
       x\pars{\expo{\alpha x} + 1}\pars{\expo{\beta x} + 1}}\,\dd x:\ {\large ?}\,,
       \qquad \alpha > \beta > 0}$.

\begin{align}
\color{#00f}{\large{\rm I}\pars{\alpha,\beta}}&=\int_{0}^{\infty}
\pars{{1 \over \expo{\beta x} + 1} - {1 \over \expo{\alpha x} + 1}}\,{\dd x \over x}
\\[3mm]&=
\int_{0}^{\infty}\braces{\half\bracks{1 - \tanh\pars{\beta x \over 2}} - \half\bracks{1 - \tanh\pars{\alpha x \over 2}}}\,{\dd x \over x}
\\[3mm]&=\lim_{\Lambda \to \infty}\braces{%
\int_{0}^{\Lambda}\half\bracks{1 - \tanh\pars{\beta x \over 2}}\,{\dd x \over x}
-
\int_{0}^{\Lambda}\half\bracks{1 - \tanh\pars{\alpha x \over 2}}\,{\dd x \over x}}
\\[3mm]&=\half\lim_{\Lambda \to\infty}\bracks{%
\int_{0}^{\Lambda}\tanh\pars{\alpha x \over 2}\,{\dd x \over x}
-\int_{0}^{\Lambda}\tanh\pars{\beta x \over 2}\,{\dd x \over x}}
\\[3mm]&=\half\lim_{\Lambda \to\infty}\bracks{%
\int_{0}^{\alpha\Lambda/2}{\tanh\pars{x} \over x}\,\dd x
-\int_{0}^{\beta\Lambda/2}{\tanh\pars{x} \over x}\,\dd x}
\\[3mm]&=\half\lim_{\Lambda \to\infty}\left\lbrace%
\bracks{\ln\pars{\alpha\Lambda \over 2}\tanh\pars{\alpha\Lambda \over 2}
-\int_{0}^{\alpha\Lambda/2}\ln\pars{x}\sech^{2}\pars{x}\,\dd x}\right.
\\[3mm]&\left.\phantom{\half\lim_{\Lambda \to\infty}\left\lbrace\right.}
\mbox{}-\bracks{\ln\pars{\beta\Lambda \over 2}\tanh\pars{\beta\Lambda \over 2}
-\int_{0}^{\beta\Lambda/2}\ln\pars{x}\sech^{2}\pars{x}\,\dd x}\right\rbrace
\\[3mm]&=\half\lim_{\Lambda \to \infty}\bracks{%
\ln\pars{\alpha\Lambda \over 2} - \ln\pars{\beta\Lambda \over 2}}
=\color{#00f}{\large\half\ln\pars{\alpha \over \beta}}
\end{align}

