Answer
$$\int x^{5}e^{-x^{3}}dx=\frac{-1}{3e^{x^{3}}}(x^{3}+1)+C$$
Work Step by Step
$$let\ t=x^{3},\,dt=3x^{2}dx$$
$$\int x^{5}e^{-x^{3}}dx=\frac{1}{3}\int te^{-t}dt$$
$$=-\frac{1}{3}te^{-t}+\frac{1}{3}\int e^{-t}dt=-\frac{1}{3}te^{-t}-\frac{1}{3}e^{-t}+C$$
$$=\frac{-1}{3e^{x^{3}}}(x^{3}+1)+C$$