<s>[INST] Math Subject GRE 1268 Question 55 If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$.   [/INST] A slight variation of the accepted solution begins on line $3$:
\begin{align}
I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\
&= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\
&= \int_{0}^{\infty} \frac{e^{-bx} dx}{(1 + e^{bx})e^{-bx}} - \int_{0}^{\infty} \frac{e^{-ax}dx}{(1 + e^{ax})e^{-ax}} \\
&=\int_{0}^{\infty} \frac{e^{-bx} dx}{(1 + e^{-bx})} - \int_{0}^{\infty} \frac{e^{-ax}dx}{(1 + e^{-ax})} \\
\end{align}
For the integral involving $b$, let $u=1+e^{-bx}$ so that $\frac{-1}{b}du=e^{-bx}dx$.  Then $x=0 \implies u=2$, and $x \to \infty \implies u \to 1^+$.  With a similar result for the integral involving $a$, and changing the order of integration, we end up with
\begin{align}
I &= \left( \frac{1}{b} - \frac{1}{a} \right) \lim _{\epsilon \to 0^+}\int_{1+\epsilon}^{2}\frac{1}{u}du = \left( \frac{1}{b} - \frac{1}{a} \right) \ln 2 \ .  \\
\end{align}   
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