From Förberedande kurs i matematik 2

With the given variable substitution, \displaystyle u=x^3, we obtain

\displaystyle du = \bigl(x^3\bigr)'\,dx = 3x^2\,dx

and because the integral contains \displaystyle x^2 as a factor, we can bundle it together with \displaystyle dx and replace the combination with \displaystyle \tfrac{1}{3}\,du,

\displaystyle \int e^{x^3}x^2\,dx = \bigl\{\,u=x^3\,\bigr\} = \int e^u\tfrac{1}{3}\,du = \frac{1}{3}e^u + C\,\textrm{.}

Thus, the answer is

\displaystyle \int e^{x^3}x^2\,dx = \frac{1}{3}e^{x^3} + C\,,

where \displaystyle C is an arbitrary constant.