Answer
$$\frac{{3{x^2} - 9}}{{{x^2} - x + 2}}$$
Work Step by Step
$$\eqalign{
& \frac{{2x - 3}}{{x - 2}} + \frac{{x + 3}}{{x + 1}} \cr
& {\text{The LCD is }}\left( {x - 2} \right)\left( {x - 1} \right),{\text{ then}} \cr
& = \frac{{\left( {2x - 3} \right)\left( {x + 1} \right)}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} + \frac{{\left( {x + 3} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} \cr
& {\text{Sum}} \cr
& = \frac{{\left( {2x - 3} \right)\left( {x + 1} \right) + \left( {x + 3} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} \cr
& {\text{Multiply and simplify}} \cr
& = \frac{{2{x^2} + 2x - 3x - 3 + {x^2} - 2x + 3x - 6}}{{{x^2} + x - 2x + 2}} \cr
& = \frac{{3{x^2} - 9}}{{{x^2} - x + 2}} \cr} $$
