Let's do the decomposition into partial fractions
(2x-1)/((x-1)^3(x-2))=A/(x-1)^3+B/(x-1)^2+C/(x-1)+D/(x-2)2x−1(x−1)3(x−2)=A(x−1)3+B(x−1)2+Cx−1+Dx−2
=(A(x-2)+B(x-2)(x-1)+C(x-2)(x-1)^2+D(x-1)^3)/((x-1)^3(x-2))=A(x−2)+B(x−2)(x−1)+C(x−2)(x−1)2+D(x−1)3(x−1)3(x−2)
Therefore,
2x-1=A(x-2)+B(x-2)(x-1)+C(x-2)(x-1)^2+D(x-1)^32x−1=A(x−2)+B(x−2)(x−1)+C(x−2)(x−1)2+D(x−1)3
Let x=2x=2, =>⇒, 3=D3=D
Let x=1x=1, =>⇒, 1=-A1=−A
Coefficients of x^3x3, =>⇒, 0=C+D0=C+D, =>⇒, C=-3C=−3
Let x=0x=0, =>⇒,-1=-2A+2B-2C-D−1=−2A+2B−2C−D
-1=2+2B+6-3−1=2+2B+6−3
2B=-1-2-3=-62B=−1−2−3=−6, =>⇒, B=-3B=−3
Therefore,
(2x-1)/((x-1)^3(x-2))=-1/(x-1)^3-3/(x-1)^2-3/(x-1)+3/(x-2)2x−1(x−1)3(x−2)=−1(x−1)3−3(x−1)2−3x−1+3x−2
