How do you integrate `x/(x^3-x^2-6x)` using partial fractions?

Quick version

Note that both the numerator and denominator are divisible by `x`, so we find:

`x/(x^3-x^2-6x) = 1/(x^2-x-6) = 1/((x-3)(x+2)) = 1/(5(x-3))-1/(5(x+2))`

So:

`int x/(x^3-x^2-6x) dx = int 1/(5(x-3))-1/(5(x+2)) dx`

`color(white)(int x/(x^3-x^2-6x) dx) = 1/5 ln abs(x-3)-1/5 ln abs(x+2) + C`

Notes

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

`color(white)(1/((x-3)(x+2))) = (A(x+2)+B(x-3))/((x-3)(x+2))`

`color(white)(1/((x-3)(x+2))) = ((A+B)x+(2A-3B))/((x-3)(x+2))`

Equating coefficients, we have:

`{ (A+B = 0), (2A-3B = 1) :}`

Adding `3` times the first equation to the second, we get:

`5A = 1`

Hence `A = 1/5` and `B=-1/5`

So:

`1/((x-3)(x+2)) = 1/(5(x-3)) - 1/(5(x+2))`

`color(white)()`
Alternatively, we could find `A` and `B` using Heaviside's cover up method:

`A = 1/(color(blue)(3)+2) = 1/5`

`B = 1/(color(blue)(-2)-3) = 1/(-5) = -1/5`

`color(white)()`
The method I actually used was:

If `1/((x-3)(x+2)) = A/(x-3)+B/(x+2)`

then when we express in terms of a common denominator we will need `A` and `B` to be the same size but opposite signs in order that the `x` term cancels out.

If try `1/(x-3)-1/(x+2)` then we get `2-(-3) = 2+3 = 5`, so that's `5` times too large.

Hence `A=1/5` and `B=-1/5`