Question: If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int\_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$. 

A: $0$

B: $1$

C: $a-b$

D: $(a-b)\log 2$

E: $\frac{a-b}{ab}\log 2$

I really don't see how to start this one, I'm not so great with integrals.

Answer: 
The integral being considered is, and is evaluated as, the following.
\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}} \\
&= \left( \frac{1}{b} - \frac{1}{a} \right) \, \int\_{1}^{\infty} \frac{dt}{t(1+t)} \mbox{ where $t = e^{bx}$ in the first and $t = e^{ax}$ in the second integral } \\
&= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim\_{p \to \infty} \, \int\_{1}^{p} \left( \frac{1}{t} - \frac{1}{1+t} \right) \, dt \\
&= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim\_{p \to \infty} \, \left[ \ln(t) - \ln(1+t) \right]\_{1}^{p} \\
&= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim\_{p \to \infty} \left[ \ln\left( \frac{p}{1 + p}\right) + \ln 2 \right] \\
&= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim\_{p \to \infty} \left[ \ln\left( \frac{1}{1 + \frac{1}{p}}\right) + \ln 2 \right] \\
&= \left( \frac{1}{b} - \frac{1}{a} \right) \, \ln 2
\end{align}

**Note:** Originally the statement "This is valid if $a \neq b$" was given at the end of the solution. Upon reflection it is believed that the statement should have been "This is valid for $a,b \neq 0$".
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}