Answer
$$\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}} + \frac{{Dx + E}}{{{x^2} + 2}}$$
Work Step by Step
$$\eqalign{
& \frac{{1 - {x^2}}}{{{x^3}\left( {{x^2} + 2} \right)}} \cr
& {x^3}{\text{ is a linear repeated factor}}{\text{, so its decomposition is }}\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}} \cr
& {x^2} {\text{ is a quadratic factor. The numerator for its decomposition is }} \cr
& {\text{a linear equation. Then}}{\text{,}} \cr
& \frac{{1 - {x^2}}}{{{x^3}\left( {{x^2} + 2} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}} + \frac{{Dx + E}}{{{x^2} + 2}} \cr} $$
