Answer
$$\frac{1}{4}{e^{{x^4}}} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^3}} {e^{{x^4}}}dx \cr
& {\text{integrate by substitution}} \cr
& {\text{set }}u = {x^4}{\text{ then }}\frac{{du}}{{dx}} = 4{x^3},\,\,\,\,\,\,\,\,{x^3}dx = \frac{1}{4}du \cr
& {\text{write the integral in terms of }}u \cr
& \int {{x^3}} {e^{{x^4}}}dx = \int {{e^u}\left( {\frac{1}{4}du} \right)} \cr
& = \frac{1}{4}\int {{e^u}du} \cr
& = \frac{1}{4}{e^u} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{4}{e^{{x^4}}} + C \cr} $$
