Answer
$$\frac{{{6^{{x^3} + 8}}}}{{3\ln 6}} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}{6^{{x^3} + 8}}dx} \cr
& {\text{substitute }}u = {x^3} + 8,{\text{ }}du = 3{x^2}dx \cr
& \int {{x^2}{6^{{x^3} + 8}}dx} = \frac{1}{3}\int {{6^u}du} \cr
& {\text{find the antiderivative}} \cr
& {\text{by the formula }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& {\text{letting }}a = 6 \cr
& = \frac{{{6^u}}}{{3\ln 6}} + C \cr
& = \frac{{{6^{{x^3} + 8}}}}{{3\ln 6}} + C \cr} $$
