We are asked to evaluate the definite integral $$ \int_{0}^{+\infty} \frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} dx $$

1. Problem Description

We are asked to evaluate the definite integral

$$$$

\int_{0}^{+\infty} \frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} dx

$$$$

2. Solution Steps

Let's rewrite the integrand:

$$$$

\frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} = \frac{e^{ax}}{(1+e^{ax})(1+e^{bx})} - \frac{e^{bx}}{(1+e^{ax})(1+e^{bx})}

$$$$

Now we rewrite the fractions further:

$$$$

\frac{e^{ax}}{(1+e^{ax})(1+e^{bx})} = \frac{1+e^{ax}-1}{(1+e^{ax})(1+e^{bx})} = \frac{1}{1+e^{bx}} - \frac{1}{(1+e^{ax})(1+e^{bx})}

$$$$

Similarly,

$$$$

\frac{e^{bx}}{(1+e^{ax})(1+e^{bx})} = \frac{1+e^{bx}-1}{(1+e^{ax})(1+e^{bx})} = \frac{1}{1+e^{ax}} - \frac{1}{(1+e^{ax})(1+e^{bx})}

$$$$

Therefore, we have

$$$$

\frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} = \left( \frac{1}{1+e^{bx}} - \frac{1}{(1+e^{ax})(1+e^{bx})} \right) - \left( \frac{1}{1+e^{ax}} - \frac{1}{(1+e^{ax})(1+e^{bx})} \right) = \frac{1}{1+e^{bx}} - \frac{1}{1+e^{ax}} = \frac{1+e^{ax} - (1+e^{bx})}{(1+e^{ax})(1+e^{bx})} = \frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})}

$$$$

Instead, we use the following decomposition:

$$$$

\frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} = \frac{1+e^{ax}-1}{1+e^{ax}} \frac{1}{1+e^{bx}} - \frac{1+e^{bx}-1}{1+e^{bx}} \frac{1}{1+e^{ax}} = \frac{1}{1+e^{bx}} - \frac{1}{1+e^{ax}}

$$$$

Thus,

$$$$

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

$$$$

$$$$

\int_0^{\infty} \frac{1}{1+e^{bx}} dx - \int_0^{\infty} \frac{1}{1+e^{ax}} dx

$$$$

$$$$

\int_0^{\infty} \frac{e^{-bx}}{1+e^{-bx}}dx - \int_0^{\infty} \frac{e^{-ax}}{1+e^{-ax}}dx

$$$$

Now, $\int \frac{e^{-cx}}{1+e^{-cx}} dx = -\frac{1}{c} \ln(1+e^{-cx})$

$$$$

-\frac{1}{b} \ln(1+e^{-bx}) \Big|_0^{\infty} - \left( -\frac{1}{a} \ln(1+e^{-ax}) \Big|_0^{\infty} \right)

$$$$

$$$$

-\frac{1}{b} [\ln(1+0) - \ln(1+1)] + \frac{1}{a} [\ln(1+0) - \ln(1+1)]

$$$$

$$$$

-\frac{1}{b} (0 - \ln 2) + \frac{1}{a} (0 - \ln 2) = \frac{\ln 2}{b} - \frac{\ln 2}{a} = \ln 2 \left(\frac{1}{b} - \frac{1}{a} \right) = \ln 2 \left(\frac{a-b}{ab}\right)

$$$$

3. Final Answer

$$$$

\ln 2 \left(\frac{a-b}{ab}\right)

$$$$