Answer
$$\frac{3}{{x - 3}} + \frac{2}{{x + 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{5x - 3}}{{{x^2} - 2x - 3}} \cr
& {\text{Factoring the denominator}} \cr
& \frac{{5x - 3}}{{\left( {x - 3} \right)\left( {x + 1} \right)}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{{5x - 3}}{{\left( {x - 3} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 3}} + \frac{B}{{x + 1}} \cr
& {\text{Multiply each side by }}\left( {x - 3} \right)\left( {x + 1} \right) \cr
& 5x - 3 = A\left( {x + 1} \right) + B\left( {x - 3} \right)\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& \cr
& {\text{Let }}x = 3{\text{ into the equation }}\left( 1 \right) \cr
& 5\left( 3 \right) - 3 = A\left( {3 + 1} \right) + B\left( {3 - 3} \right) \cr
& 12 = A\left( 4 \right) \cr
& A = 3 \cr
& {\text{Let }}x = - 1{\text{ into the equation }}\left( 1 \right) \cr
& 5\left( { - 1} \right) - 3 = A\left( { - 1 + 1} \right) + B\left( { - 1 - 3} \right) \cr
& - 8 = B\left( { - 4} \right) \cr
& B = 2 \cr
& \cr
& {\text{Use }}A{\text{ and }}B{\text{ to find the partial fraction decomposition}} \cr
& \frac{{5x - 3}}{{\left( {x - 3} \right)\left( {x + 1} \right)}} = \frac{3}{{x - 3}} + \frac{2}{{x + 1}} \cr} $$
