# How do you use partial fraction decomposition to decompose the fraction to integrate (x^2-x-8)/((x+1)(x^2+5x+6))?

Aug 13, 2015

$\frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)} = - \frac{3}{x + 1} + \frac{2}{x + 2} + \frac{2}{x + 3}$

#### Explanation:

$\frac{{x}^{2} - x - 8}{\left(x + 1\right) \left({x}^{2} + 5 x + 6\right)}$

First we need to finish factoring the denominator:

$\frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)}$

Now we want $A , B , \text{and } C$ to get:

$\frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3} = \frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)}$

So we need:

$A \left(x + 2\right) \left(x + 3\right) + B \left(x + 1\right) \left(x + 3\right) + C \left(x + 1\right) \left(x + 2\right) = {x}^{2} - x - 8$

The left side can be rewritten:

$A {x}^{2} + 5 A x + 6 A + B {x}^{2} + 4 B x + 3 B + C {x}^{2} + 3 C x + 2 C$

and then:

$\left(A + B + C\right) {x}^{2} + \left(5 A + 4 B + 3 C\right) x + \left(6 A + 3 B + 2 C\right) = {x}^{2} - x - 8$

So we need to solve the system:

$A + B + C = 1$

$5 A + 4 B + 3 C = - 1$

$6 A + 3 B + 2 C = - 8$

Multiplying the first equation by $- 3$ and adding to the second. Then the first times $- 2$ and add to the third, gets us:

$2 A + B = - 4$

$4 A + B = - 10$

So $2 A = - 6$ and $A = - 3$, which gets us $B = 2$ and together these give us $C = 2$.

$\frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)} = - \frac{3}{x + 1} + \frac{2}{x + 2} + \frac{2}{x + 3}$