Answer
$$\frac{1}{{x + 2}} - \frac{1}{{x - 2}} + \frac{2}{{{x^2} + 4}}$$
Work Step by Step
$$\eqalign{
& \frac{{ - 2{x^2} - 24}}{{{x^4} - 16}} \cr
& {\text{Factor the denominator}} \cr
& \frac{{ - 2{x^2} - 24}}{{{x^4} - 16}} = \frac{{ - 2{x^2} - 24}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {{x^2} + 4} \right)}} \cr
& {\text{Decompose }}\frac{{ - 2{x^2} - 24}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {{x^2} + 4} \right)}}{\text{ into partial fractions}} \cr
& \frac{{ - 2{x^2} - 24}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {{x^2} + 4} \right)}} = \frac{A}{{x + 2}} + \frac{B}{{x - 2}} + \frac{{Cx + D}}{{{x^2} + 4}} \cr
& {\text{Multiply each side by }}\left( {x + 2} \right)\left( {x - 2} \right)\left( {{x^2} + 4} \right) \cr
& - 2{x^2} - 24 = A\left( {x - 2} \right)\left( {{x^2} + 4} \right) + B\left( {x + 2} \right)\left( {{x^2} + 4} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {Cx + D} \right)\left( {{x^2} - 4} \right) \cr
& {\text{Expand and combine like terms on the right}} \cr
& - 2{x^2} - 24 = A{x^3} - 2A{x^2} + 4Ax - 8A + B{x^3} + 2B{x^2} + 4Bx + 8B \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + C{x^3} + D{x^2} - 4Cx - 4D \cr
& - 2{x^2} - 24 = \left( {A + B\, + C} \right){x^3} + \left( { - 2A + 2B + D} \right){x^2} + \left( {4A + 4B - 4C} \right)x \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 8A + 8B - 4D \cr
& {\text{Equating the coefficients}} \cr
& A + B\, + C = 0 \cr
& - 2A + 2B + D = - 2 \cr
& 4A + 4B - 4C = 0 \cr
& \,\, - 8A + 8B - 4D = - 24 \cr
& {\text{Solving the system of equations using a CAS we obtain}} \cr
& A = 1,\,\,\,B = - 1,\,\,\,C = 0,\,\,\,\,D = 2,\,\, \cr
& {\text{Use }}A,{\text{ }}B,\,C,\,\,{\text{ and }}D{\text{ to find the partial fraction decomposition}} \cr
& \frac{{ - 2{x^2} - 24}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {{x^2} + 4} \right)}} = \frac{1}{{x + 2}} - \frac{1}{{x - 2}} + \frac{2}{{{x^2} + 4}} \cr} $$
