Question Detail

This integral requires the method of partial fractions.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 by the common denominator $(x-2)(x+1)$:$4x - 5 = A(x+1) + B(x-2)$To find A, set $x=2$:$4(2) - 5 = A(2+1) \implies 8 - 5 = 3A \implies 3 = 3A \implies A = 1$To find B, set $x=-1$:$4(-1) - 5 = B(-1-2) \implies -4 - 5 = -3B \implies -9 = -3B \implies B = 3$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 = \ln|x-2| + C_1$$\int \frac{3}{x+1} dx = 3\ln|x+1| + C_2$Combining these, the result is:$\ln|x-2| + 3\ln|x+1| + C$