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