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: 
Tale $a = 1$ and let $b\to 0^+.$ In the limit you get

$$\int\_0^\infty\frac{e^x-1}{(1+e^x)2}\,dx.$$

That integral equals $\infty$ because the integrand has a positive limit at $\infty.$ The only answer that fits this phenomenon is E.
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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> $$
> \color{blue}{\int\_{0}^{\infty}\frac{e^{bx}-e^{ax}}{\left(1+e^{bx}\right)\left(1+e^{ax}\right)}dx=\left(\frac{1}{b}-\frac{1}{a}\right)\log\left(2\right).}$$
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