# How do you express 1/(x^3-4x) in partial fractions?

$\frac{1}{{x}^{3} - 4 x} = \frac{1}{x \left(x - 2\right) \left(x + 2\right)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}$
Now try various values of $x$ except $0 , - 2 , 2$ to find the values of
constants $A , B , C$ finally you get
$\frac{1}{{x}^{3} - 4 x} = - \frac{1}{4 x} + \frac{1}{8 \cdot \left(x - 2\right)} + \frac{1}{8 \cdot \left(x + 2\right)}$