Answer
$$ - \frac{1}{{4\left( {x + 1} \right)}} + \frac{1}{{4\left( {x - 1} \right)}} + \frac{1}{{2\left( {{x^2} + 1} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{{{x^4} - 1}} \cr
& {\text{Factor the denominator}} \cr
& \frac{{{x^2}}}{{{x^4} - 1}} = \frac{{{x^2}}}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} \cr
& {\text{Decompose }}\frac{{{x^2}}}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{{{x^2}}}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{x + 1}} + \frac{B}{{x - 1}} + \frac{{Cx + D}}{{{x^2} + 1}} \cr
& {\text{Multiply each side by }}\left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^2} + 1} \right) \cr
& {x^2} = A\left( {x - 1} \right)\left( {{x^2} + 1} \right) + B\left( {x + 1} \right)\left( {{x^2} + 1} \right) + \left( {Cx + D} \right)\left( {{x^2} - 1} \right) \cr
& {\text{Expand and combine like terms on the right}} \cr
& {x^2} = \left( {A + B + C} \right){x^3} + \left( { - A + B + D} \right){x^2} + \left( {A + B - C} \right)x \cr
& \,\,\,\,\,\,\,\,\, - A + B - D \cr
& {\text{Equating the coefficients}} \cr
& A + B + C = 0 \cr
& - A + B + D = 1 \cr
& A + B - C = 0 \cr
& - A + B - D = 0 \cr
& {\text{Solving the system of equations using a CAS we obtain}} \cr
& A = - \frac{1}{4},\,\,\,B = \frac{1}{4},\,\,\,C = 0,\,\,\,\,D = \frac{1}{2},\,\, \cr
& {\text{Use }}A,{\text{ }}B,\,C,\,\,{\text{ and }}D{\text{ to find the partial fraction decomposition}} \cr
& \frac{{{x^2}}}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{{ - 1/4}}{{x + 1}} + \frac{{1/4}}{{x - 1}} + \frac{{1/2}}{{{x^2} + 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \frac{1}{{4\left( {x + 1} \right)}} + \frac{1}{{4\left( {x - 1} \right)}} + \frac{1}{{2\left( {{x^2} + 1} \right)}} \cr} $$
