[tex]\dfrac{x^3-9x}{x+3} =\dfrac{x(x^2-9)}{x+3} =\dfrac{x.(x-3).(x+3)}{x+3} =x(x-3)=x^2-3x\\ \\ \\ \\\int\limits^{-1}_{0} {\dfrac{x^3-9x}{x+3}} \, dx =\int\limits^{-1}_{0} {x^2-3x} \, dx=\dfrac{x^3}{3}-\dfrac{3x^2}{2} =0-(-\dfrac{1}{3}-\dfrac{3}{2})=\dfrac{11}{6}[/tex]
