If $a,b>0$, find $\int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$

Question

The question is to find $$\int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$$ where $a$ and $b$ are positive values. My attempt was to break this up into the sum $$ \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx=\int_0^\infty \frac{1}{(1+e^{ax})}dx-\int_0^\infty \frac{1}{(1+e^{bx})}dx $$ and then calculate each individual integral, but I am not sure how to complete the question. It seems like $u$ sub may work, but it was not following easily so I am not sure if this is the best way. Also, as this was taken from a Math GRE practice exam, I am thinking that there may be a trick to this that I am not seeing, so any insight would be appreciated.

Answer 1

Let $u=e^{ax}$. Then $du=ae^{ax}\,dx$, so our first integral is $$\int_1^{\infty}\frac{1}{au(u+1)}\,du.$$ We are integrating $\frac{1}{a}\left(\frac{1}{u}-\frac{1}{u+1}\right)$. One antiderivative is $\frac{1}{a}\ln\left(\frac{u}{u+1}\right)$.