Answer
$$\frac{2}{x} + \frac{3}{{x + 2}} - \frac{1}{{x - 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{4{x^2} - 3x - 4}}{{{x^3} + {x^2} - 2x}} \cr
& {\text{Factor the denominator}} \cr
& \frac{{4{x^2} - 3x - 4}}{{{x^3} + {x^2} - 2x}} = \frac{{4{x^2} - 3x - 4}}{{x\left( {x + 2} \right)\left( {x - 1} \right)}} \cr
& {\text{Decompose }}\frac{{4{x^2} - 3x - 4}}{{x\left( {x + 2} \right)\left( {x - 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{{4{x^2} - 3x - 4}}{{x\left( {x + 2} \right)\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{x - 1}} \cr
& \cr
& {\text{Multiply each side by }}x\left( {x + 2} \right)\left( {x - 1} \right) \cr
& 4{x^2} - 3x - 4 = A\left( {x + 2} \right)\left( {x - 1} \right) + Bx\left( {x - 1} \right) + Cx\left( {x + 2} \right)\,\,\left( {\bf{1}} \right) \cr
& {\text{Let }}x = 0{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& 4{\left( 0 \right)^2} - 3\left( 0 \right) - 4 = A\left( 2 \right)\left( { - 1} \right) + B\left( 0 \right) + C\left( 0 \right)\, \cr
& - 4 = - 2A \to A = 2 \cr
& {\text{Let }}x = - 2{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& 4{\left( { - 2} \right)^2} - 3\left( { - 2} \right) - 4 = A\left( 0 \right) + B\left( { - 2} \right)\left( { - 2 - 1} \right) + C\left( 0 \right)\,\, \cr
& 18 = 6B \to B = 3 \cr
& {\text{Let }}x = 1{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& 4{\left( 1 \right)^2} - 3\left( 1 \right) - 4 = A\left( 0 \right) + B\left( 0 \right) + C\left( 1 \right)\left( 3 \right)\,\, \cr
& - 3 = C3 \to C = - 1 \cr
& {\text{Use }}A,{\text{ }}B,\,\,\,{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{{4{x^2} - 3x - 4}}{{x\left( {x + 2} \right)\left( {x - 1} \right)}} = \frac{2}{x} + \frac{3}{{x + 2}} - \frac{1}{{x - 1}} \cr} $$
