Question

How do you use partial fraction decomposition to decompose the fraction to integrate `(x^2-x-6)/(x^3 +3x)`?

Answer 1

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

Explanation:

`(x^2-x-6)/(x^3 +3x)`

First we need to factor the denominator into quadratic and linear polynomials that are irreducible using Real coefficients.

`x^3+3x = x(x^2+3)`
Neither `x` nor `x^2 + 3` can be further factored with Real coefficients.

So we need to find `A, B, "and " C` to make:

`A/x +(Bx+C)/(x^2+3) = (x^2-x-6)/(x^3 +3x)`

Clear the denominators or get a common denominator on the left to see that we need:

`Ax^2 +3A +Bx^2+Cx = x^2 - x -6`

`(A+B)x^2 +(C)x+(3A) = 1x^2 -1x-6`

Setting the coefficients equal to each other, we need to solve:

`A+B=1`
`C = -1`
`3A = -6`

We can quickly see that we need `A = -2` and `C = -1`, so the first equation becomes:
`-2+B = 1`, so `B =3`

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