Question Detail

To simplify this integral, we use partial fraction decomposition.First, factor the denominator: $x^2-x-2 = (x-2)(x+1)$.Next, set up the partial fraction decomposition:$\frac{4x-5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$Multiply both sides by $(x-2)(x+1)$:$4x-5 = A(x+1) + B(x-2)$To find $A$, set $x=2$:$4(2)-5 = A(2+1) + B(0) \implies 3 = 3A \implies A=1$To find $B$, set $x=-1$:$4(-1)-5 = A(0) + B(-1-2) \implies -9 = -3B \implies B=3$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 (with terms reordered).