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

##### 1 Answer
Aug 22, 2015

$\frac{{x}^{2} - x - 6}{{x}^{3} + 3 x} = \frac{- 2}{x} + \frac{3 x - 1}{{x}^{2} + 3}$

#### Explanation:

$\frac{{x}^{2} - x - 6}{{x}^{3} + 3 x}$

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

${x}^{3} + 3 x = x \left({x}^{2} + 3\right)$
Neither $x$ nor ${x}^{2} + 3$ can be further factored with Real coefficients.

So we need to find $A , B , \text{and } C$ to make:

$\frac{A}{x} + \frac{B x + C}{{x}^{2} + 3} = \frac{{x}^{2} - x - 6}{{x}^{3} + 3 x}$

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

$A {x}^{2} + 3 A + B {x}^{2} + C x = {x}^{2} - x - 6$

$\left(A + B\right) {x}^{2} + \left(C\right) x + \left(3 A\right) = 1 {x}^{2} - 1 x - 6$

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

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

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

$\frac{A}{x} + \frac{B x + C}{{x}^{2} + 3} = \frac{- 2}{x} + \frac{3 x - 1}{{x}^{2} + 3}$