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 $a,b \neq 0$".
Another approach. $$\int\_{0}^{\infty}\frac{e^{-bx}}{1+e^{-bx}}dx=\int\_{0}^{\infty}\sum\_{k\geq1}\left(-1\right)^{k+1}e^{-kbx}dx=\sum\_{k\geq1}\left(-1\right)^{k+1}\int\_{0}^{\infty}e^{-kbx}dx
 $$ $$=\frac{1}{b}\sum\_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}=\frac{1}{b}\log\left(2\right)
 $$ and in similar way we can calculate $\int\_{0}^{\infty}\frac{e^{-ax}}{1+e^{-ax}}dx
 $. So 

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