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: 
*One indirect approach:*

Write 

$$f(a,b) = \int\_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$$

Then changing variables by $x = ku$, for some positive $k$,

$$f(a,b) = k \int\_0^\infty \frac{e^{kau}-e^{kbu}}{(1+e^{kau})(1+e^{kbu})}du = k\, f(ka,kb) $$

The only$^\*$ answer which obeys the relation $$f(a,b) = k \,f(ka,kb)$$ is Option E. 

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*$^\*$Footnote:* As pointed out in the comments, Option A does follow this relation as well but is easy to rule out on other grounds. 

To unpack my original thinking:

* Option A is not possible as we can make the integrand positive: for $a > b$, $f(a,b) > 0$
* Option B is not possible as $f(a,b)$ cannot be independent of $a$ and $b$, e.g., without explicitly calculating, it looks clear that $\partial\_a f(a,b) \neq 0$.
* Now we're down to Options C, D or E. Since this is a question from a timed test and mathematicians are ruthlessly efficient (aka lazy), I don't want to evaluate the integral. Instead, can I find quickly some sort of scaling argument to rule out Options C and D? My 'fear' is that $f(ka,kb) = k\,f(a,b)$ which will instead rule out Option E and then I'll be stuck having to calculate the integral in order to differentiate between C and D.
* But no! Instead $f(a,b) = k\,f(ka,kb)$. Good!
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 

> 
> $$
> \color{blue}{\int\_{0}^{\infty}\frac{e^{bx}-e^{ax}}{\left(1+e^{bx}\right)\left(1+e^{ax}\right)}dx=\left(\frac{1}{b}-\frac{1}{a}\right)\log\left(2\right).}$$
> 
> 
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