Question

How do you integrate `7/(x-3)^2` using partial fractions?

Answer 1

`(-7)/(x-3)+c`

Explanation:

`7/(x-3)^2=((A1)/(x-3))+((A2)/(x-3)^2)`
`A1*(x-3)+A2=7`
For x = 3,
`A2=7`
Comparing coefficient of x, we get `A=0`
Let `t=(x-3), dt=dx`
`int(7/t^2)dt=(-7/t)+c`
Substituting value of `t=(x-3)`
`=(-7)/(x-3)+c`