Answer
$$\frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}}$$
Work Step by Step
$$\eqalign{
& \frac{5}{{x\left( {{x^2} - 4} \right)}} \cr
& {\text{Factor the difference of two squares }}{x^2} - 4 \cr
& \frac{5}{{x\left( {{x^2} - 4} \right)}} = \frac{5}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} \cr
& {\text{All the factors are in the denominator,}}{\text { so the numerator for each factor }} \cr
& {\text{is a constant}}{\text{.}} \cr
& {\text{The partial fraction decomposition is in the form:}} \cr
& \frac{5}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}} \cr
& \frac{5}{{x\left( {{x^2} - 4} \right)}} = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}} \cr} $$
