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.
Clearly (A) and (B) are wrong. Noting $$\lim\_{a\to\infty} f(a,b) = \lim\_{a\to\infty}\int\_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx=\int\_0^\infty \frac{1}{1+e^{bx}}dx<\infty,$$
we have to choose (E).