Answer
$$ - \frac{1}{{2{x^2}}} + \frac{1}{{2\left( {{x^2} - 2} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{1}{{{x^2}\left( {{x^2} - 2} \right)}} \cr
& \frac{1}{{{x^2}\left( {{x^2} - 2} \right)}} = \frac{1}{{{x^2}\left( {x + 2} \right)\left( {x - 2} \right)}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{1}{{{x^2}\left( {{x^2} - 2} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{{Cx + D}}{{{x^2} - 2}} \cr
& {\text{Multiply each side by }}{x^2}\left( {x + 2} \right)\left( {x - 2} \right) \cr
& 1 = Ax\left( {{x^2} - 2} \right) + B\left( {{x^2} - 2} \right) + \left( {Cx + D} \right){x^2}\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& {\text{Expand and combine like terms on the right}} \cr
& 1 = A{x^3} - 2Ax + B{x^2} - 2B + C{x^3} + D{x^2}\, \cr
& 1 = A{x^3} + C{x^3} + B{x^2} + D{x^2} - 2Ax - 2B\, \cr
& {\text{Equating the coefficients}} \cr
& - 2B = 1,\,\,\,\,B = - \frac{1}{2} \cr
& - 2A = 0,\,\,\,A = 0 \cr
& B + D = 0,\,\,\,\,D = \frac{1}{2} \cr
& A + C = 0,\,\,\,\,C = 0 \cr
& {\text{Use }}A,{\text{ }}B,\,\,C{\text{ and }}D{\text{ to find the partial fraction decomposition}} \cr
& \frac{1}{{{x^2}\left( {{x^2} - 2} \right)}} = \frac{0}{x} + \frac{{ - 1/2}}{{{x^2}}} + \frac{{1/2}}{{{x^2} - 2}} \cr
& \frac{1}{{{x^2}\left( {{x^2} - 2} \right)}} = - \frac{1}{{2{x^2}}} + \frac{1}{{2\left( {{x^2} - 2} \right)}} \cr} $$
