Question

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

Answer 1

`int x/(x-3 )^2dx = ln(|x-3|) - 3/(x-3) + C`, where `C` is the constant of integration.

Explanation:

Using the partial fractions decomposition, we write `x/(x-3 )^2` in the form of `A/(x-3) + B/(x-3)^2`, which is easier to integrate.

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

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

`iff A = 1 and -3A + B = 0 iff A = 1 and B = 3.`

Therefore, `x/(x-3 )^2 = 1/(x-3) + 3/(x-3)^2.`

And `int x/(x-3 )^2dx = int 1/(x-3) dx + 3* int 1/(x-3)^2 dx`

`int x/(x-3 )^2dx = ln(|x-3|) - 3/(x-3) + C`, where `C` is the constant of integration.