Integral of (3x-2)/(x-1)^2

1 answer:

7 0

(3x−2)/(x−1)^2=A/(x−1) + B/x(x−1)^2
 =[A(x−1)+Bx(x−1)] / 2 3x-2=A(x-1)+Bx
3x-2=x(A+B)-A
A+B=3
-A=-2=>A=2
A+B=3=>2+B=3=>B=1

lets check our partial fraction we have 2/(x−1) + x/(x−1)^2 = [2(x−1)+x] / (x−1)^2
 =(3x−2) / (x−1)^2

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