Answer
$$ - \frac{1}{{12}}{e^{ - 3{x^4}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{{e^{3{x^4}}}}}} dx \cr
& {\text{Use property of exponents}} \cr
& = \int {{e^{ - 3{x^4}}}{x^3}} dx \cr
& {\text{set }}u = - 3{x^4}{\text{ then }}\frac{{du}}{{dx}} = - 12{x^3},\,\,\,\,\,\, - \frac{1}{{12}}du = {x^3}dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{e^{ - 3{x^4}}}{x^3}} dx = \int {{e^u}} \left( { - \frac{1}{{12}}du} \right) \cr
& = - \frac{1}{{12}}\int {{e^u}} du \cr
& {\text{integrating by the formula }}\int {{e^u}} du = {e^u} + C \cr
& = - \frac{1}{{12}}{e^u} + C \cr
& {\text{replace }} - 3{x^4}{\text{ for }}u \cr
& = - \frac{1}{{12}}{e^{ - 3{x^4}}} + C \cr} $$
