Question Detail

First, factor the denominator: ( x^2-x-2 = (x-2)(x+1) ).Use partial fraction decomposition:( \frac{4x-5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} )Multiply 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) \Rightarrow 3 = 3A \Rightarrow A=1 ).To find ( B ), 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 )( = \ln|x-2| + 3\ln|x+1| + C ).This matches option B.