Evaluate: ∫ (3e²ˣ - 2eˣ)/(e²ˣ + 2eˣ - 8) dx

Put $e^x = t$, so $e^x\,dx = dt$ and $dx = \frac{dt}{t}$. $\int \frac{3e^{2x} - 2e^x}{e^{2x}+2e^x-8}dx = \int \frac{3t-2}{t^2+2t-8}dt = \int \frac{3t-2}{(t+4)(t-2)}dt$ Partial fractions: $\frac{3t-2}{(t+4)(t-2)} = \frac{A}{t+4} + \frac{B}{t-2}$, giving $A = \frac{7}{3}$, $B = \frac{2}{3}$. $= \frac{7}{3}\log|t+4| + \frac{2}{3}\log|t-2| + C$ Hence the result is $\frac{7}{3}\log(e^x+4) + \frac{2}{3}\log|e^x-2| + C$.