Answer
$$\frac{{1/5}}{{x - 4}} + \frac{{4/5}}{{x + 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{{x^2} - 3x}}{{{x^3} - 3{x^2} - 4x}} \cr
& {\text{factoring}} \cr
& = \frac{{x\left( {x - 3} \right)}}{{x\left( {{x^2} - 3x - 4} \right)}} = \frac{{x - 3}}{{{x^2} - 3x - 4}} \cr
& = \frac{{x - 3}}{{\left( {x - 4} \right)\left( {x + 1} \right)}} \cr
& {\text{partial fraction decomposition}} \cr
& \frac{{x - 3}}{{\left( {x - 4} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 1}} \cr
& x - 3 = A\left( {x + 1} \right) + B\left( {x - 4} \right) \cr
& {\text{for }}x = 4 \cr
& 4 - 3 = A\left( {4 + 1} \right) + B\left( {4 - 4} \right) \cr
& 1 = 5A \cr
& 1/5 = A \cr
& {\text{for }}x = - 1 \cr
& - 1 - 3 = A\left( { - 1 + 1} \right) + B\left( { - 1 - 4} \right) \cr
& - 4 = - 5B \cr
& 4/5 = B \cr
& {\text{substituting the values}} \cr
& \frac{A}{{x - 4}} + \frac{B}{{x + 1}} = \frac{{1/5}}{{x - 4}} + \frac{{4/5}}{{x + 1}} \cr} $$
