Question

What is the partial-fraction decomposition of `(x^2+2x+7)/(x(x-1)^2)`?

Answer 1

`(x^2+2x+7)/(x(x-1)^2) = 7/x-6/(x-1)+10/(x-1)^2`

Explanation:

The factors on the denominator are obvious. They are all linear, but one of them is a double factor. So we want `A, B, and C` so that:

`(x^2+2x+7)/(x(x-1)^2) = A/x+B/(x-1)+C/(x-1)^2`

Combining the rations on the right, we get a numerator pf:
`A(x^2-2x+1) +B(x^2-x) +Cx = Ax^2-2Ax+A+Bx^2-Bx+Cx`

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

Setting the coefficients equal to those of the original numerator, `x^2+2x+7`, we get:

`A+B=1`
`-2A-B+C = 2`
`A=7`

It is immediate that `A=7` and from that and the first equation (the coefficients of `x^2`), we get `B=-6`. Substituting in the middle equation and solving for `C`, we get `C=10`.

`(x^2+2x+7)/(x(x-1)^2) = 7/x-6/(x-1)+10/(x-1)^2`

It is a good idea to check the answer by getting the common denominator. (I did that on paper, but I'm not going to type it up.)