Evaluating an indefinite integral with exponents and logarithms I was taking a GRE practice exam and came across 
$$ \int_0^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1 + e^{bx})} dx $$
I noted that this can be expressed as 
$$ \int_0^{\infty} \frac{1}{(1 + e^{bx})} - \frac{1}{(1 + e^{ax})} dx $$
And 
$$ \int \frac{1}{(1 + e^{cx})} dx  , u = e^{cx}, du = ce^{cx}  dx \rightarrow$$
$$ \frac{1}{c} \int \frac{1}{u(1+u)} du = \frac{1}{c} \left(\ln(u) - \ln(1 + u) \right)  = \frac{1}{c} \ln \left( \frac{e^{cx}}{1 + e^{cx}} \right)$$ 
So then 
$$ \int_0^{\infty} \frac{1}{(1 + e^{bx})} - \frac{1}{(1 + e^{ax})} dx = \frac{1}{b} \ln \left( \frac{e^{bx}}{1 + e^{bx}} \right) - \frac{1}{a} \ln \left( \frac{e^{ax}}{1 + e^{ax}} \right) | _0^{\infty} $$ 
Which yields that we are attempting to evaluate
$$ \ln \left( \frac{(1 + e^{ax})^{\frac{1}{a}}}{(1 + e^{bx})^{\frac{1}{b}}} \right) |_0^{\infty} $$ 
Which reduces to evaluating:
$$ \frac{(1 + e^{ax})^{\frac{1}{a}}}{(1 + e^{bx})^{\frac{1}{b}}}  |_0^{\infty} $$
Which is:
$$ \lim_{x \rightarrow \infty} \frac{(1 + e^{ax})^{\frac{1}{a}}}{(1 + e^{bx})^{\frac{1}{b}}}  - 1 $$ 
But at this point I can't seem to crack this with L'hopital's rule. And in general, this problem should be completed between 30-seconds to 1 minute so I think this entire approach is invalid since its taking longer than that.

Hint:
$$
\ln \left(\frac{e^{cx}}{1+e^{cx}}\right)\Big|_{x=0}^{x=\infty} = \ln 1 -  \ln \frac{1}{2}
$$
