How do you express #(2x - 1) / [(x - 1)^3 (x - 2)]# in partial fractions?

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Answer 1 · 2017-07-14T12:51:41

The answer is #=-1/(x-1)^3-3/(x-1)^2-3/(x-1)+3/(x-2)#

Let's perform 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)#

#=(A(x-2)+B(x-1)(x-2)+C(x-1)^2(x-2)+D(x-1)^3)/((x-1)^3(x-2))#

The denominators are the same, we compare the numerators

#2x-1=A(x-2)+B(x-1)(x-2)+C(x-1)^2(x-2))+D(x-1)^3#

Let #x=1#, #=>#, #1=-A#, #=>#, #A=-1#

Let #x=2#, #=>#, #3=D#

Coefficients of #x^3#

#0=C+D#, #=>#, #C=-D=-3#

Let #x=0#, #-1=-2A+2B-2C-D#

#2B=2A+2C+D-1=-2-6+3-1=-6#

#B=-3#

Therefore,

#(2x-1)/((x-1)^3(x-2))=-1/(x-1)^3-3/(x-1)^2-3/(x-1)+3/(x-2)#