How do you find #int (x^2 - 4) / (x -1)^3 dx# using partial fractions?

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Answer 1 · 2017-12-27T07:29:36

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

Perform the decomposition into partial fractions

#(x^2-4)/(x-1)^3=A/(x-1)^3+B/(x-1)^2+C/(x-1)#

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

The denominators are the same, compare the numerators

#x^2-4=A+B(x-1)+C(x-1)^2#

Let #x=1#, #=>#, #-3=A#

Coefficient of #x^2#

#1=C#

Coefficients of #x#

#0=B-2C#, #=>#, #B=2C=2#

So,

#(x^2-4)/(x-1)^3=-3/(x-1)^3+2/(x-1)^2+1/(x-1)#

#int((x^2-4)dx)/(x-1)^3=-3intdx/(x-1)^3+2intdx/(x-1)^2+intdx/(x-1)#

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