Answer
$$\frac{1}{3}{x^3}{e^{{x^3}}} - \frac{1}{3}{e^{{x^3}}} + C $$
Work Step by Step
$$\eqalign{
& \int {{x^5}} {e^{{x^3}}}dx \cr
& = \int {{x^3}} {e^{{x^3}}}{x^2}dx \cr
& \,\,\,\,\,{\text{Let }}t = {x^3},\,\,\,\,dt = 3{x^2}dx, \cr
& \int {{x^3}} {e^{{x^3}}}{x^2}dx = \int t {e^t}\left( {\frac{1}{3}dt} \right) \cr
& = \frac{1}{3}\int t {e^t}dt \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = t,\,\,\,\,du = dt,\,\,\,\,dv = {e^t}dt,\,\,\,\,v = {e^t} \cr
& {\text{Integration by parts then gives}} \cr
& \frac{1}{3}\int t {e^t}dt = \frac{1}{3}\left( {t{e^t} - \int {{e^t}} dt} \right) \cr
& \frac{1}{3}\int t {e^t}dt = \frac{1}{3}t{e^t} - \frac{1}{3}\int {{e^t}} dt \cr
& \frac{1}{3}\int t {e^t}dt = \frac{1}{3}t{e^t} - \frac{1}{3}{e^t} + C \cr
& \cr
& {\text{write in terms of }}x{\text{: substitute }}{x^3}{\text{ for }}t \cr
& \int {{x^5}} {e^{{x^3}}}dx = \frac{1}{3}{x^3}{e^{{x^3}}} - \frac{1}{3}{e^{{x^3}}} + C \cr} $$
