Question

How do you integrate `int (x-x^2)/((x+3)(x-1)(x+4))` using partial fractions?

Answer 1

`int (x-x^2)/((x+3)(x-1)(x+4)) dx = 3 ln abs(x+3)-4 ln abs(x+4) + C`

Explanation:

`(x-x^2)/((x+3)(x-1)(x+4)) = A/(x+3)+B/(x-1)+C/(x+4)`

We can find `A`, `B` and `C` using Heaviside's cover up method:

`A=((color(blue)(-3))-(color(blue)(-3))^2)/(((color(blue)(-3))-1)((color(blue)(-3))+4)) = (-12)/((-4)(1)) = 3`

`B=((color(blue)(1))-(color(blue)(1))^2)/(((color(blue)(1))+3)((color(blue)(1))+4)) = 0/((4)(5)) = 0`

`C=((color(blue)(-4))-(color(blue)(-4))^2)/(((color(blue)(-4))+3)((color(blue)(-4))-1)) = (-20)/((-1)(-5)) = -4`

So:

`int (x-x^2)/((x+3)(x-1)(x+4)) dx = int 3/(x+3)-4/(x+4) dx`

`color(white)(int (x-x^2)/((x+3)(x-1)(x+4)) dx) = 3 ln abs(x+3)-4 ln abs(x+4) + C`