Question

How do you integrate `(3x) / (x-3)^2` using partial fractions?

Answer 1

The answer is `=-9/(x-1)+3ln(x-3)+C`

Explanation:

Let's start the decomposition in partial fractions
`(3x)/(x-3)^2=A/(x-3)^2+B/(x-3)`

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

`:.` `3x=A+B(x-3)`
coefficients of x, `B=3`
`0=A-3B` `=>` `A=9`

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

`int(3xdx)/(x-3)^2=int(9dx)/(x-3)^2+int(3dx)/(x-3)`

`=-9/(x-1)+3ln(x-3)+C`