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.
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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.
orion
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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 $\frac 1 {1+e^{bx}}-\frac 1 {1+e^{ax}}$.
Kavi Rama Murthy
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