Answer
$$5{x^2} + \frac{3}{x} - \frac{1}{{x + 3}} + \frac{2}{{x - 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{5{x^5} + 10{x^4} - 15{x^3} + 4{x^2} + 13x - 9}}{{{x^3} + 2{x^2} - 3x}} \cr
& {\text{By using the long division}} \cr
& = 5{x^2} + \frac{{4{x^2} + 13x - 9}}{{{x^3} + 2{x^2} - 3x}} \cr
& {\text{Decompose }}\frac{{4{x^2} + 13x - 9}}{{{x^3} + 2{x^2} - 3x}}{\text{ into partial fractions}} \cr
& \frac{{4{x^2} + 13x - 9}}{{{x^3} + 2{x^2} - 3x}} = \frac{{4{x^2} + 13x - 9}}{{x\left( {x + 3} \right)\left( {x - 1} \right)}} \cr
& \frac{{4{x^2} + 13x - 9}}{{x\left( {x + 3} \right)\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x + 3}} + \frac{C}{{x - 1}} \cr
& \cr
& {\text{Multiply each side by }}x\left( {x + 3} \right)\left( {x - 1} \right) \cr
& 4{x^2} + 13x - 9 = A\left( {x + 3} \right)\left( {x - 1} \right) + Bx\left( {x - 1} \right) + Cx\left( {x + 3} \right)\,\,\left( {\bf{1}} \right) \cr
& {\text{Let }}x = 0{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& - 9 = A\left( {0 + 3} \right)\left( {0 - 1} \right) + B\left( 0 \right) + C\left( 0 \right) \cr
& - 9 = - 3A \to A = 3 \cr
& {\text{Let }}x = - 3{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& 4{\left( { - 3} \right)^2} + 13\left( { - 3} \right) - 9 = A\left( 0 \right) + B\left( { - 3} \right)\left( { - 3 - 1} \right) + C\left( 0 \right)\,\, \cr
& - 12 = 12B \to B = - 1 \cr
& {\text{Let }}x = 1{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& 4{\left( 1 \right)^2} + 13\left( 1 \right) - 9 = A\left( 0 \right) + B\left( 0 \right) + C\left( 1 \right)\left( {1 + 3} \right)\,\, \cr
& 8 = C\left( 1 \right)\left( 4 \right)\, \to C = 2 \cr
& {\text{Use }}A,{\text{ }}B,\,\,\,{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{{4{x^2} + 13x - 9}}{{x\left( {x + 3} \right)\left( {x - 1} \right)}} = \frac{3}{x} - \frac{1}{{x + 3}} + \frac{2}{{x - 1}} \cr
& \cr
& {\text{Therefore,}} \cr
& 5{x^2} + \frac{{4{x^2} + 13x - 9}}{{{x^3} + 2{x^2} - 3x}} = 5{x^2} + \frac{3}{x} - \frac{1}{{x + 3}} + \frac{2}{{x - 1}} \cr} $$
