Question

How do you express `(6x^2+1)/(x^2(x-1)^2)` in partial fractions?

Answer 1

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

Explanation:

`(6x^2+1)/(x^2(x-1)^2) = A/x+B/x^2+C/(x-1)+D/(x-1)^2`

So:

`6x^2+1 = Ax(x-1)^2+B(x-1)^2+Cx^2(x-1)+Dx^2`

Putting `x=1` we get:

`7 = D`

Putting `x=0` we get:

`1 = B`

So:

`6x^2+1 = Ax(x-1)^2+(x-1)^2+Cx^2(x-1)+7x^2`

`color(white)(6x^2+1) = Ax(x^2-2x+1)+(x^2-2x+1)+Cx^2(x-1)+7x^2`

`color(white)(6x^2+1) = A(x^3-2x^2+x)+(x^2-2x+1)+C(x^3-x^2)+7x^2`

`color(white)(6x^2+1) = (A+C)x^3+(8-2A-C)x^2+(A-2)x+1`

Hence:

`A=2" "` and `" "C = -2`

So:

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