Question Detail

First, factor the denominator: $x^2-x-2 = (x-2)(x+1)$.Now, use partial fraction decomposition:$\frac{4x-5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$Multiply both sides by $(x-2)(x+1)$: $4x-5 = A(x+1) + B(x-2)$.Set $x=2$: $4(2)-5 = A(2+1) \Rightarrow 3 = 3A \Rightarrow A = 1$.Set $x=-1$: $4(-1)-5 = B(-1-2) \Rightarrow -9 = -3B \Rightarrow B = 3$.So the integral becomes:$\int \left( \frac{1}{x-2} + \frac{3}{x+1} \right) dx$Integrate each term:$= \int \frac{1}{x-2} dx + \int \frac{3}{x+1} dx = \ln|x-2| + 3\ln|x+1| + C$.This matches option B.