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

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Answer 1 · 2016-06-01T14:57:22

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

The expansion for #x^2/(x-1)^3# is given by

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

reducing the right member to common denominator we have

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

Equating the numerators we have

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

giving the following conditions

#{ (A=1), (B-2A=0), (A-B+C=0) :}#

solving for #A,B,C# we get

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