Question

How do you integrate `[6/(x-3)^4*(x^2-4x+4)]` using partial fractions?

Answer 1

See the answer below:

Answer 2

`(-6x^2+30x-38)/(x-3)^3+C`

Explanation:

You don't. Instead, substitute `x=u+3`, `dx=du` and the integral becomes `6 int 1/u^2+3/u^3+1/u^4du`
`=6(-1/u-1/u^2-(1/3)/u^3)+C` leading, upon substituting `u=x-3`, to the answer. This substitution always works well if the denominator is simply a power of a linear expression.