#### Answer

$$\frac{{{{10}^{{x^3}}}}}{{3\ln 10}} + C$$

#### Work Step by Step

$$\eqalign{
& \int {{x^2}{{10}^{{x^3}}}} dx \cr
& {\text{substitute }}u = {x^3},{\text{ }}du = 3{x^2}dx \cr
& \int {{x^2}{{10}^{{x^3}}}dx} = \frac{1}{3}\int {{{10}^u}du} \cr
& {\text{find the antiderivative}} \cr
& {\text{by the formula }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& {\text{letting }}a = 10 \cr
& = \frac{{{{10}^u}}}{{3\ln 10}} + C \cr
& = \frac{{{{10}^{{x^3}}}}}{{3\ln 10}} + C \cr} $$