Answer
$${x^3} - {x^2} - \frac{1}{{2\left( {2x + 1} \right)}} + \frac{2}{{3\left( {x + 2} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{{2{x^5} + 3{x^4} - 3{x^3} - 2{x^2} + x}}{{2{x^2} + 5x + 2}} \cr
& {\text{Using the long division}} \cr
& {x^3} - {x^2} + \frac{x}{{2{x^2} + 5x + 2}} \cr
& {\text{Decompose }}\frac{x}{{2{x^2} + 5x + 2}}{\text{ into partial fractions}} \cr
& \frac{x}{{2{x^2} + 5x + 2}} = \frac{x}{{\left( {2x + 1} \right)\left( {x + 2} \right)}} \cr
& \frac{x}{{\left( {2x + 1} \right)\left( {x + 2} \right)}} = \frac{A}{{2x + 1}} + \frac{B}{{x + 2}} \cr
& x = A\left( {x + 2} \right) + B\left( {2x + 1} \right) \cr
& {\text{If }}x = - 1/2 \cr
& - \frac{1}{2} = A\left( {\frac{3}{2}} \right) \cr
& A = - \frac{1}{2} \cr
& {\text{If }}x = - 2 \cr
& - 2 = A\left( 0 \right) + B\left( { - 3} \right) \cr
& B = \frac{2}{3} \cr
& {\text{Use }}A,{\text{ and }}B{\text{ to find the partial fraction decomposition}} \cr
& \frac{x}{{\left( {2x + 1} \right)\left( {x + 2} \right)}} = - \frac{1}{{2\left( {2x + 1} \right)}} + \frac{2}{{3\left( {x + 2} \right)}} \cr
& \frac{{2{x^5} + 3{x^4} - 3{x^3} - 2{x^2} + x}}{{2{x^2} + 5x + 2}} = {x^3} - {x^2} - \frac{1}{{2\left( {2x + 1} \right)}} + \frac{2}{{3\left( {x + 2} \right)}} \cr} $$
