Answer
$$1 + \frac{1}{x} + \frac{5}{{x - 2}} - \frac{3}{{x - 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{{x^3} + 2}}{{{x^3} - 3{x^2} + 2x}} \cr
& {\text{Use the long division}} \cr
& \frac{{{x^3} + 2}}{{{x^3} - 3{x^2} + 2x}} = 1 + \frac{{3{x^2} - 2x + 2}}{{{x^3} - 3{x^2} + 2x}} \cr
& {\text{Decompose }}\frac{{3{x^2} - 2x + 2}}{{{x^3} - 3{x^2} + 2x}} \cr
& \frac{{3{x^2} - 2x + 2}}{{{x^3} - 3{x^2} + 2x}} = \frac{{3{x^2} - 2x + 2}}{{x\left( {{x^2} - 3x + 2} \right)}} = \frac{{3{x^2} - 2x + 2}}{{x\left( {x - 2} \right)\left( {x - 1} \right)}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{{3{x^2} - 2x + 2}}{{x\left( {x - 2} \right)\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 2}} + \frac{C}{{x - 1}} \cr
& {\text{Multiply each side by }}x\left( {x - 2} \right)\left( {x - 1} \right) \cr
& 3{x^2} - 2x + 2 = A\left( {x - 2} \right)\left( {x - 1} \right) + Bx\left( {x - 1} \right) + Cx\left( {x - 2} \right)\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& \cr
& {\text{Let }}x = 0{\text{ into the equation }}\left( 1 \right) \cr
& 2 = A\left( { - 2} \right)\left( { - 1} \right) + B\left( 0 \right) + C\left( 0 \right)\, \cr
& A = 1 \cr
& \cr
& {\text{Let }}x = 2{\text{ into the equation }}\left( 1 \right) \cr
& 10 = A\left( 0 \right) + 2B\left( {2 - 1} \right) + C\left( 0 \right)\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& B = 5 \cr
& \cr
& {\text{Let }}x = 1{\text{ into the equation }}\left( 1 \right) \cr
& 3 = A\left( 0 \right) + B\left( 0 \right) + C\left( {1 - 2} \right)\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& C = - 3 \cr
& \cr
& {\text{Use }}A,{\text{ }}B{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{{3{x^2} - 2x + 2}}{{x\left( {x - 2} \right)\left( {x - 1} \right)}} = \frac{1}{x} + \frac{5}{{x - 2}} - \frac{3}{{x - 1}} \cr
& \frac{{{x^3} + 2}}{{{x^3} - 3{x^2} + 2x}} = 1 + \frac{1}{x} + \frac{5}{{x - 2}} - \frac{3}{{x - 1}} \cr} $$
