Question

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

Answer 1

`I=3ln(x-3) −9/(x−3) +C`

Explanation:

To integrate `I = 3int (x)/(x^2-6x+9) dx` we will use Partial Fraction Decomposition of the form
`(x)/(x^2-6x+9) = A/(x-3) + B/(x-3)^2`
`(x)/cancel(x^2-6x+9) = (A(x-3) + B)/cancel(x^2-6x+9)` Now determine A and B setting `x=0` and `x=1`
`x= 0: => 0 = -3A + B`; `=> 3A = B`
`x= 1: => 1 = -2A + B` substitute the above and
`1=-2A + 3A; A= 1` and substituting into above `B = 3`
So now our integral becomes:
`I =3int (1/(x-3) + 3/(x-3)^2) dx` Applying linearity
`I = 3(I_1 + I_2) = 3(int 1/(x-3)dx + int 3/(x-3)^2dx)`
`I_1 = int 1/(x-3)dx = ln(x-3)` you can use substitution here
`I_2 = 3int 1/(x-3)^2dx = −3/(x−3)` again use substitution method and apply power rule:
`I = 3(I_1 + I_2)=3(ln(x-3) −3/(x−3))` the final answer need a constant of course:
`I= 3ln(x-3) −9/(x−3) +C`