Question

How do you express `(3x) / (x-3)^2` in partial fractions?

Answer 1

`(3x)/(x-3)^2hArr3/(x-3)+9/(x-3)^2`

Explanation:

Partial fractions of `(3x)/(x-3)^2` are in the form `A/(x-3)+B/(x-3)^2`

i.e. `(3x)/(x-3)^2hArrA/(x-3)+B/(x-3)^2`

or `(3x)/(x-3)^2hArr(A(x-3)+B)/(x-3)^2=(Ax-3A+B)/(x-3)^2`

Comparing like terms `A=3` and `-3A+B=0` i.e. `B=3A=9`

Hence `(3x)/(x-3)^2hArr3/(x-3)+9/(x-3)^2`