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

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Answer 1 · 2016-12-20T14:19:53

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

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)#

#=(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)^3#

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

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

Coefficients of #x^3#, #=>#, #0=C+D#, #=>#, #C=-3#

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

#-1=2+2B+6-3#

#2B=-1-2-3=-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)#