Answer
$$\frac{1}{3}{e^{{x^3} + 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}{e^{{x^3} + 1}}} dx \cr
& {\text{Integrate by using the substitution method}} \cr
& u = {x^3} + 1,{\text{ }}du = 3{x^2}dx,{\text{ }}dx = \frac{1}{{3{x^2}}}du \cr
& \int {{x^2}{e^{{x^3} + 1}}} dx = \int {{x^2}{e^{{x^3} + 1}}} \left( {\frac{1}{{3{x^2}}}} \right)du \cr
& = \int {{e^{{x^3} + 1}}} \left( {\frac{1}{3}} \right)du \cr
& = \frac{1}{3}\int {{e^u}du} \cr
& = \frac{1}{3}{e^u} + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{3}{e^{{x^3} + 1}} + C \cr} $$
