# How do you express (2x+ 3)/(x^4-9x^2) in partial fractions?

$\frac{2 x + 3}{{x}^{4} - 9 {x}^{2}} = - \frac{2}{9 x} + \frac{1}{3 {x}^{2}} - \frac{1}{18 \left(x + 3\right)} - \frac{1}{6 \left(x - 3\right)}$
Looking to the denominator ${x}^{4} - 9 {x}^{2} = {x}^{2} \left(x + 3\right) \left(x - 3\right)$ then the fraction admits an expansion such that
$\frac{2 x + 3}{{x}^{4} - 9 {x}^{2}} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x + 3} + \frac{D}{x - 3}$ so
doing $f = \frac{2 x + 3}{{x}^{4} - 9 {x}^{2}} - \frac{A}{x} - \frac{B}{x} ^ 2 - \frac{C}{x + 3} - \frac{D}{x - 3} = 0$ and calculanting $f \left({x}^{4} - 9 {x}^{2}\right) = \left(D + C - A\right) {x}^{3} + \left(3 D + B - 3 C\right) {x}^{2} + \left(9 A + 2\right) x + 3 - 9 B = 0$ for all values of x, we get:
$A = - \frac{2}{9} , B = \frac{1}{3} , C = - \frac{1}{18} , D = - \frac{1}{6}$