Integral of (3x-2)/(x-1)^2
(3x−2)/(x−1)^2=A/(x−1) + B/x(x−1)^2 =[A(x−1)+Bx(x−1)] / 23x-2=A(x-1)+Bx3x-2=x(A+B)-AA+B=3-A=-2=>A=2A+B=3=>2+B=3=>B=1lets check our partial fractionwe have2/(x−1) + x/(x−1)^2 = [2(x−1)+x] / (x−1)^2 =(3x−2) / (x−1)^2
