Question

How do you integrate `int 1/(x^3-6x^2+9x)` using partial fractions?

Answer 1

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

Explanation:

The integrand may be written `1/(x(x^2-6x+9))`
`-=1/(x(x-3)^2)`
By using the cover-up rule for partial fractions, putting `x=0`, to give the `1/9`:
`-=(1/9)(1/x)+(Ax+B)/(x-3)^2` with `A` and `B` to be found.
Multiplying through by `x(x^2-6x+9)` :
`1-=(1/9)(x-3)^2+Ax^2+Bx` so
`0x^2+0x+1-=(1/9+A)x^2+(-2/3+B)x+1`
For this is identity to be true the coefficients of the powers of `x` must be the same on the two sides. So `A=-1/9`, `B=2/3`.
This gives the integral as
`1/9int1/x+(6-x)/(x-3)^2dx`

`=1/9ln|x|+1/9int(6-x)/(x-3)^2dx`
By substituting `u=x-3`, `dx=du`, or otherwise, the second integral becomes `-3/(x-3)-ln|x-3|` leading to the answer.