Answer
$$\frac{1}{3}\left( {{v^3} - {e^{3v}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {{v^2} - {e^{3v}}} \right)} dv \cr
& {\text{split integrand}} \cr
& = \int {{v^2}} dv - \int {{e^{3v}}} dv \cr
& {\text{use integration rules }}\int {{v^n}} dv = \frac{{{v^{n + 1}}}}{{n + 1}} + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {{e^{av}}dv} = \frac{1}{a}{e^{av}} + C \cr
& = \frac{{{v^3}}}{3} - \frac{1}{3}\left( {{e^{3v}}} \right) + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{3}{v^3} - \frac{1}{3}{e^{3v}} + C \cr
& {\text{factor}} \cr
& = \frac{1}{3}\left( {{v^3} - {e^{3v}}} \right) + C \cr} $$