Evaluating an indefinite integral with exponents and logarithms

Question

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.

Answer 1

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

Answer 2

Another way to evaluate the integral, $$\int_0^\infty \frac{dx}{1+e^{ax}}$$ which I think is a little faster (so it might help on the GRE). Is to multiplty the integrand by $\frac{e^{-ax}}{e^{-ax}}$ to get $$\int_0^\infty \frac{e^{-ax}}{1+e^{-ax}}dx.$$ Which you can either use a usub (with $u=1+e^{-ax})$ or just see that it is a natural logarithm to get, $$\frac{-1}{a}ln|1+e^{-ax}| |_0^\infty = \frac{ln(2)}{a}.$$