Question

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

Answer 1

`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`