This is from GRE Math subject test, Question #55, from https://www.ets.org/s/gre/pdf/practice_book_math.pdf
If $a,b > 0$, then what is the value of $$ \int_0^\infty \frac {e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx $$
I'm not sure if it's Calc II integration or Residue Theorem from complex analysis, but I have no idea where to start ...
One may write $$ \frac {e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}=\frac {(1+e^{ax})-(1+e^{bx})}{(1+e^{ax})(1+e^{bx})}=\frac 1{1+e^{bx}}-\frac 1{1+e^{ax}} $$ then, by the change of variable $u=e^{ax}$, $x=\dfrac1a \:\ln u$, $dx=\dfrac1a \:\dfrac{du}u$, one gets $$ \int_0^\infty\frac {dx}{1+e^{ax}}=\dfrac1a\int_1^\infty\frac {du}{u(1+u)}=\dfrac1a\ln 2 $$ giving $$ \int_0^\infty \frac {e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx=\left(\frac1b-\dfrac1a\right)\cdot \ln 2=\frac{a-b}{ab}\cdot \ln 2. $$