Answer
$\frac{1}{3}e^{x^3}+C$
Work Step by Step
Let $u=x^3$. Then $du=3x^2 dx$, and $\frac{1}{3}du=x^2dx$.
$\int x^2 e^{x^3}dx$
$=\int e^{x^3}x^2 dx$
$=\int e^u*\frac{1}{3}du$
$=\frac{1}{3}e^u+C$
$=\frac{1}{3}e^{x^3}+C$
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