Factorization to solve improper integral with exponentials

Question

How calculate $$\int_{0}^{\infty} \dfrac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$$ with $a$, $b$ are positive values? I think that there is some algebraic manipulation that I can not see.

Answer 1

Partial fractions, the integrand is just

$$\frac{1}{1+e^{bx}}-\frac{1}{1+e^{ax}}$$

which can then be easily integrated by taking the denominator as a new variable in each term.

Answer 2

The integrand is nothing but $\frac {1+e^{ax}} {(1+e^{ax})(1+e^{bx})}-\frac {1+e^{bx}} {(1+e^{ax})(1+e^{bx})}$ which is $\frac 1 {1+e^{bx}}-\frac 1 {1+e^{ax}}$.