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

Dec 14, 2015

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

#### Explanation:

Factor the denominator.

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

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

If $x = - 2$:

$8 = - 24 A$
$A = - \frac{1}{3}$

If $x = 2$:

$8 = 24 B$
$B = \frac{1}{3}$

If $x = 0$:

$4 = - 4 A + 4 B - 4 D$
$1 = - A + B - D$

Plug in $A$ and $B$ to see that $D = - \frac{1}{3}$.

If $x = 1$:

$5 = - 3 A + 9 B - 3 C - 3 D$

Plug in all the known values to find that $C = 0$.

Plug these all back in to see that the expression decomposes into

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