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