Question

How do you use partial fraction decomposition to decompose the fraction to integrate `(x+2)/(x^3-2x^2+x)`?

Answer 1

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

Explanation:

Factor the denominator:

`x^3-2x^2+x = x(x^2-2x+1) = x(x-1)^2`

Find `A, B, " and " C` to solve:

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

That is:

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

(When I was learning this is was clearest to me to get a common denominator on the left. So I'll show it that way.)

`(A(x-1)^2)/(x(x-1)^2) +(Bx(x-1))/(x(x-1)^2) +(Cx)/(x(x-1)^2) = (x+2)/(x(x-1)^2)`

`(Ax^2-2Ax+A+Bx^2-Bx+Cx)/(x(x-1)^2) = (x+2)/(x(x-1)^2)`

Regrouping and setting the numerators equal to each other, we get:

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

`= 0x^2+1x+2`

So

`A+B=0`
`-2A-B+C=1`
`A=2`

From the first and last equation we can see that: `A=2` and `B=-2`.

Substituting in the middle equation, we find that `C=3`

So:

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