Use Heaviside's method to find \(A\) and \(B\).
\[ \dfrac{-2x+9}{(x+3)(x-2)}\cdot (x+3)=\dfrac{-2x+9}{x-2} \] \[ A= \dfrac{-2x+9}{x-2)}\Huge|_{\normalsize x=-3}\normalsize=-3 \] \[ \dfrac{-2x+9}{(x+3)(x-2)}\cdot (x-2)=\dfrac{-2x+9}{x+3} \] \[ B= \dfrac{-2x+9}{x+3)}\Huge|_{\normalsize x=2}\normalsize=1 \]Therefore,
\[ \dfrac{x^2-x+3}{x^2+x-6}=1-\dfrac{3}{x+3}+\dfrac{1}{x-2} \]