Question

How do you integrate `int x/((x-6)(x-3) dx` using partial fractions?

Answer 1

`intx/((x-6)(x-3))dx = 2ln|x-6| - ln|x-3| + C`

Explanation:

To find the partial fraction decomposition, we begin with

`x/((x-6)(x-3)) = A/(x-6) + B/(x-3)`

Multiplying by `(x-6)(x-3)` gives us

`x = A(x-3)+B(x-6) = (A+B)x + (-3A - 6B)`

Matching respective coefficients on each side gives us the system of equations

`{(A + B = 1), (-3A -6B = 0):}`

Now all we need to do is solve for `A` and `B`. For example, using elimination,

`(-3A - 6B) + 3(A+B) = 0 + 3(1)`
`=> -3B = 3`
`=> B = -1`

Then, substituting

`A - 1 = 1`
`=> A = 2`

So the decomposition is

`x/((x-6)(x-3)) = 2/(x-6) - 1/(x-3)`

Now we can do the integral relatively easily.

`intx/((x-6)(x-3))dx = int(2/(x-6) - 1/(x-3))dx`

`int(2/(x-6) - 1/(x-3))dx = 2int(1/(x-6))dx - int(1/(x-3))dx`

A couple of simple `u` substitutions, together with `int(1/x)dx = ln|x|+C`
gives us

`2int(1/(x-6))dx = 2ln|x-6| + C`
and
`-int(1/(x-3))dx = -ln|x-3|+C`

So putting it all together, we get

`intx/((x-6)(x-3))dx = 2ln|x-6| - ln|x-3| + C`