How do you use partial fraction decomposition to integrate #(−10x^2+27x−14)/((x−1)^3(x+2))#?

1 Answer
Monzur R.
Oct 21, 2017

#int(−10x^2+27x−14)/((x−1)^3(x+2))dx=#
#int(4/(x+2)-4/(x-1)+2/(x-1)^2+1/(x-1)^3)dx#

Explanation:

Since the question is asking about how to use partial fractions to perform the integration, I shall only provide the partial fraction decomposition and leave the rest to the questioner.

#(−10x^2+27x−14)/((x−1)^3(x+2))=#
#A/(x+2)+B/(x-1)+C/(x-1)^2+D/(x-1)^3#

#−10x^2+27x−14=#
#A(x-1)^3+B(x-1)^2(x+2)+C(x-1)(x+2)+D(x+2)#

Let #x=2#

Then #3=3DrArrD=1#

Let #x=-2#

Then #-108=-27ArArrA=4#

Notice that if we had expanded the RHS, we would've found two cubic terms with coefficients #A# and #B#. And since #LHS=RHS,(A+B)x^3=x^3 rArrA+B=0rArr4+B=0rArrB=-4#

Finally letting #x=0#, we get

#-14=A+2B-2C+2D#

By simplifying and subbing in the values we already have, we get #C=2#

#therefore(−10x^2+27x−14)/((x−1)^3(x+2))=#
#4/(x+2)-4/(x-1)+2/(x-1)^2+1/(x-1)^3#

Thus #int(−10x^2+27x−14)/((x−1)^3(x+2))dx=#
#int(4/(x+2)-4/(x-1)+2/(x-1)^2+1/(x-1)^3)dx#

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