Answer
$-\displaystyle \frac{1}{4}e^{(-x^{4}+8)}+C$
Work Step by Step
$\displaystyle \int\frac{x^{3}}{e^{(x^{4}-8)}}dx=\int x^{3}\cdot e^{-(x^{4}-8)}dx=\int x^{3}\cdot e^{-x^{4}+8}dx$
Shortcut to apply:
$\displaystyle \quad \int g\cdot e^{u}dx=\frac{g}{u'}e^{u}+C$
$\left[\begin{array}{ll}
g(x)=x^{3}, & u(x)=-x^{4}+8,\\
& u'(x)=-4x^{3}
\end{array}\right]$
$=\displaystyle \frac{x^{3}}{-4x^{3}}e^{-x^{4}+8}+C$
= $-\displaystyle \frac{1}{4}e^{(-x^{4}+8)}+C$
