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

1. Problem Description

We need to evaluate the definite integral:

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

2. Solution Steps

First, let's simplify the expression by multiplying both the numerator and denominator by $e^{-(a+b)x}$:

$\frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1 + e^{bx})} = \frac{e^{ax} - e^{bx}}{1 + e^{ax} + e^{bx} + e^{(a+b)x}} = \frac{e^{-bx} - e^{-ax}}{(e^{-ax} + 1)(e^{-bx} + 1)} e^{-(a+b)x} \cdot e^{(a+b)x} = \frac{e^{-bx}-e^{-ax}}{1 + e^{-ax}+e^{-bx}+e^{-(a+b)x}} e^{-(a+b)x} e^{(a+b)x}$

Multiplying numerator and denominator by $e^{-(a+b)x}$:

$\frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} = \frac{e^{ax} - e^{bx}}{1 + e^{ax} + e^{bx} + e^{(a+b)x}} \cdot \frac{e^{-(a+b)x}}{e^{-(a+b)x}} = \frac{e^{-bx} - e^{-ax}}{(1 + e^{-ax})(1 + e^{-bx})}$

Let $u=e^{-ax}$ and $v=e^{-bx}$

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

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

Now, we divide the numerator and the denominator of the first integral by $e^{ax}$:

$\int_{0}^{\infty} \frac{1}{(1 + e^{ax})(e^{-ax} + e^{(b-a)x})}dx = \int_{0}^{\infty} \frac{1}{e^{-ax}+1+e^{(b-a)x}+e^{bx}}$

Divide the numerator and denominator by $e^{ax}$

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

$\frac{e^{-bx} - e^{-ax}}{1 + e^{-bx} + e^{-ax} + e^{-(a+b)x}}$

Let $I = \int_{0}^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1 + e^{bx})}dx = \int_{0}^{\infty} \frac{e^{ax} - e^{bx}}{1 + e^{ax} + e^{bx} + e^{(a+b)x}} dx$.

We can rewrite it as:

$I = \int_{0}^{\infty} \frac{e^{ax}}{1 + e^{ax} + e^{bx} + e^{(a+b)x}} dx - \int_{0}^{\infty} \frac{e^{bx}}{1 + e^{ax} + e^{bx} + e^{(a+b)x}} dx$

$= \int_{0}^{\infty} \frac{e^{ax}}{1 + e^{ax}} \frac{1}{1 + e^{bx}} dx - \int_{0}^{\infty} \frac{e^{bx}}{1 + e^{bx}} \frac{1}{1 + e^{ax}} dx$.

We can also rewrite the integral as:

$I = \int_{0}^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1 + e^{bx})}dx = \int_{0}^{\infty} \frac{e^{ax} - e^{bx}}{1 + e^{ax} + e^{bx} + e^{(a+b)x}} dx$

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

$= \int_{0}^{\infty} \frac{1}{1 + e^{bx}}dx - \int_{0}^{\infty} \frac{1}{1 + e^{ax}}dx = \int_{0}^{\infty} \frac{1}{1 + e^{bx}} - \frac{1}{1 + e^{ax}} dx$

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

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

$= [-\frac{1}{b} \ln(e^{-bx} + 1)]_0^{\infty} - [-\frac{1}{a} \ln(e^{-ax} + 1)]_0^{\infty} = [-\frac{1}{b} \ln(0 + 1) + \frac{1}{b} \ln(1 + 1)] - [-\frac{1}{a} \ln(0 + 1) + \frac{1}{a} \ln(1 + 1)]$

$= \frac{\ln 2}{b} - \frac{\ln 2}{a} = \ln 2 (\frac{1}{b} - \frac{1}{a}) = \ln 2 (\frac{a - b}{ab})$

But, notice that the integrand is anti-symmetric under the exchange of a and b.

So, $I = \int_{0}^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1 + e^{bx})} dx = \int_{0}^{\infty} \frac{e^{ax}}{ (1 + e^{ax})(1 + e^{bx})} dx - \int_{0}^{\infty} \frac{e^{bx}}{(1 + e^{ax})(1 + e^{bx})} dx$

$= \frac{1}{b} ln(\frac{2}{1}) - \frac{1}{a}ln(\frac{2}{1})$

Then:

$\frac{1}{a}ln|1 + e^{0}| - \frac{1}{a}lim_{x \to \infty}ln|1 + e^{-ax}| =$ ln 2/a

Final Answer: The final answer is $\ln(2)\left(\frac{a-b}{ab}\right)$

3. Final Answer

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