Answer
$ - \frac{1}{4}{e^{ - {t^4}}} + C$
Work Step by Step
$$\eqalign{
& \int {{t^3}{e^{ - {t^4}}}} dt \cr
& {\text{Let }}u = - {t^4},{\text{ then }}du = - 4{t^3}dt,{\text{ }}{t^3}dt = - \frac{1}{4}du \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int {{t^3}{e^{ - {t^4}}}} dt = \int {{e^{ - {t^4}}}} \overbrace {{t^3}dt}^{ - \frac{1}{4}du} = \int {{e^u}\left( { - \frac{1}{4}} \right)} du \cr
& = - \frac{1}{4}\int {{e^u}} du \cr
& {\text{Integrating}} \cr
& = - \frac{1}{4}{e^u} + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}u = - {t^4} \cr
& = - \frac{1}{4}{e^{ - {t^4}}} + C \cr} $$
