Question

How do you integrate `int (3x^3 - 22x^2 + 27x + 46)/(x^2 - 8x + 15)` using partial fractions?

Answer 1

`int (3x^3-22x^2+27x+46)/(x^2-8x+15) dx = 3/2x^2+2x-5ln abs(x-3)+3ln abs(x-5) + C`

Explanation:

`(3x^3-22x^2+27x+46)/(x^2-8x+15) = ((3x^3-24x^2+45x)+(2x^2-16x+30)+(-2x+16))/(x^2-8x+15)`

`color(white)((3x^3-22x^2+27x+46)/(x^2-8x+15)) = 3x+2+(-2x+16)/(x^2-8x+15)`

`color(white)((3x^3-22x^2+27x+46)/(x^2-8x+15)) = 3x+2+(-2x+16)/((x-3)(x-5))`

`color(white)((3x^3-22x^2+27x+46)/(x^2-8x+15)) = 3x+2+A/(x-3)+B/(x-5)`

We can determine `A` and `B` using Heaviside's cover up method...

`A = (-2(color(blue)(3))+16)/(color(blue)(3)-5) = 10/(-2) = -5`

`B = (-2(color(blue)(5))+16)/(color(blue)(5)-3) = 6/2 = 3`

So:

`int (3x^3-22x^2+27x+46)/(x^2-8x+15) dx = int 3x+2-5/(x-3)+3/(x-5) dx`

`color(white)(int (3x^3-22x^2+27x+46)/(x^2-8x+15) dx) = 3/2x^2+2x-5ln abs(x-3)+3ln abs(x-5) + C`