91Ó°ÊÓ

Here we are given the definite integral: \(\int_{0}^{4} \frac{6}{1+e^{0.5 x}} d x\)

First, let's make the substitution: \(u = 1 + e^{0.5x}\), and so \(du = 0.5e^{0.5x} dx\), which leads to \(2 du = e^{0.5x} dx\). But we know that \(e^{0.5x} = u - 1\), so the substitution becomes \(2 du = (u - 1) dx\), meaning that \(dx = 2 du / (u - 1)\). After substituting this back into the integral, we get `\(\int \frac{6 (2 du / (u - 1))}{u}\), that simplifies to the integral \(\int \frac{12 du}{u (u - 1)}\). This can be integrated by partial fractions, separating it into two different integrals: \(\int \frac{12 du}{u}\) and \(\int {-12 du}{u - 1}\). These integrals yield \(12 ln |u|\) and \(-12 ln |u - 1|\) respectively. Therefore the antiderivative of our original function is \(12 ln |1 + e^{0.5x}|\) - \(12 ln |1 + e^{0.5x} - 1|\).

Now we substitute the lower and upper limits of the definite integral. We plug in \(x = 4\) and \(x = 0\) into our expression and subtract the second result from the first one. This gives us \(12 ln (1 + e^2) - 12 ln e^2\) - \(12 ln 2 - 12 ln 1 = 12 (ln (1 + e^2) - ln e^2 - ln 2 + ln 1)\). As \(ln 1 = 0\), we simplify to get \(12 (ln (1 + e^2) - 2 ln e - ln 2)\), which is the final answer.