Answer
\[\frac{1}{3}{e^{{x^3} - 3x}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,\left( {{x^2} - 1} \right){e^{{x^3} - 3x}}dx} \hfill \\
Let\,\,u = {x^3} - 3x \hfill \\
So\,that\,\,du = 3{x^2} - 3 \hfill \\
\,\,\,\,\,\,du = 3\,\left( {{x^2} - 1} \right)dx \hfill \\
Then \hfill \\
\frac{1}{3}\int_{}^{} {3\,\left( {{x^2} - 1} \right){e^{{x^3} - 3x}}dx} \hfill \\
\frac{1}{3}\int_{}^{} {{e^u}du} \hfill \\
Integrating\,\, \hfill \\
\frac{1}{3}{e^u} + C \hfill \\
Substituting\,\,u = {x^3} - 3x\,\,gives \hfill \\
\frac{1}{3}{e^{{x^3} - 3x}} + C \hfill \\
\end{gathered} \]
