# How do you write the partial fraction decomposition of the rational expression  (x^2+9)/(x^4-2x^2-8)?

Dec 18, 2015

$- \frac{13}{24 \left(x + 2\right)} + \frac{13}{24 \left(x - 2\right)} - \frac{7}{6 \left({x}^{2} + 2\right)}$

#### Explanation:

Factor the denominator.

${x}^{4} - 2 {x}^{2} - 8 = \left({x}^{2} - 4\right) \left({x}^{2} + 2\right) = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2\right)$

$\frac{{x}^{2} + 9}{\left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2\right)} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{{x}^{2} + 2}$

Find a common denominator of $\left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2\right)$.

${x}^{2} + 9 = A \left(x - 2\right) \left({x}^{2} + 2\right) + B \left(x + 2\right) \left({x}^{2} + 2\right) + \left(C x + D\right) \left({x}^{2} - 4\right)$

${x}^{2} + 9 = A {x}^{3} - 2 A {x}^{2} + 2 A x - 4 A + B {x}^{3} + 2 B {x}^{2} + 2 B x + 4 B + C {x}^{3} - 4 C x + D {x}^{2} - 4 D$

${x}^{2} + 9 = {x}^{3} \left(A + B + C\right) + {x}^{2} \left(- 2 A + 2 B + D\right) + x \left(2 A + 2 B + 4 C\right) + 1 \left(- 4 A + 4 B - 4 D\right)$

From this, write the following system:
$\left\{\begin{matrix}A + B + C = 0 \\ - 2 A + 2 B + D = 1 \\ 2 A + 2 B + 4 C = 0 \\ - 4 A + 4 B - 4 D = 9\end{matrix}\right.$

Solve the system:
$\left\{\begin{matrix}A = - \frac{13}{24} \\ B = \frac{13}{24} \\ C = 0 \\ D = - \frac{7}{6}\end{matrix}\right.$

This gives:

$\frac{{x}^{2} + 9}{{x}^{4} - 2 {x}^{2} - 8} = - \frac{13}{24 \left(x + 2\right)} + \frac{13}{24 \left(x - 2\right)} - \frac{7}{6 \left({x}^{2} + 2\right)}$