Question

How do you integrate `(x^2-x-8)/((x+1)(x^2+5x+6))` using partial fractions?

Answer 1

`int (x^2-x-8)/((x+1)(x+2)(x+3))`

`= -3ln abs(x+1)+2ln abs(x+2)+2ln abs(x+3) + C`

Explanation:

Note that:

`x^2+5x+6 = (x+2)(x+3)`

So:

`(x^2-x-8)/((x+1)(x+2)(x+3)) = A/(x+1)+B/(x+2)+C/(x+3)`

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

`A = ((color(blue)(-1))^2-(color(blue)(-1))-8)/(((color(blue)(-1))+2)((color(blue)(-1))+3)) = (-6)/2 = -3`

`B = ((color(blue)(-2))^2-(color(blue)(-2))-8)/(((color(blue)(-2))+1)((color(blue)(-2))+3)) = (-2)/(-1) = 2`

`C = ((color(blue)(-3))^2-(color(blue)(-3))-8)/(((color(blue)(-3))+1)((color(blue)(-3))+2)) = 4/2 = 2`

So:

`int (x^2-x-8)/((x+1)(x+2)(x+3))`

`= int (-3)/(x+1)+2/(x+2)+2/(x+3) color(white)(.)dx`

`= -3ln abs(x+1)+2ln abs(x+2)+2ln abs(x+3) + C`

Answer 2

`int \ (x^2-x-8)/((x+1)(x^2+5x+6)) \ dx = -3ln|x+1| + 2ln|x+2| + 2ln|x+3| + c`

Explanation:

We seek:

`I = int \ (x^2-x-8)/((x+1)(x^2+5x+6)) \ dx`
`\ \ = int \ (x^2-x-8)/((x+1)(x+2)(x+3)) \ dx`

First let us decompose the fraction into partial fractions, which will be of the form:

`(x^2-x-8)/((x+1)(x+2)(x+3)) -= A/(x+1) + B/(x+2) + C/(x+3)`
`" " = (A(x+2)(x+3) + B(x+1)(x+3) + C(x+2)(x+1)) / ((x+1)(x+2)(x+3))`

Leading to the identity:

`x^2-x-8 -= A(x+2)(x+3) + B(x+1)(x+3) + C(x+2)(x+1)`

Where `A,B` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x = -1 => 1+1-8 = A(1)(2) => A = -3`
Put `x = -2 => 4+2-8 = B(-1)(1) => B = 2`
Put `x = -3 => 9+3-8 = C(-1)(-2) => C = 2`

Thus we have:

`I = int \ -3/(x+1) + 2/(x+2) + 2/(x+3) \ dx`

Which, we can now integrate to get:

`I = -3ln|x+1| + 2ln|x+2| + 2ln|x+3| + c`