Answer
\[\frac{1}{2}{e^{2{x^3}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {3{x^2}{e^{2{x^3}}}dx} \hfill \\
Let\,\,u = 2{x^3} \hfill \\
So\,\,that\,\,du = 6{x^2}dx.\,\,Then \hfill \\
\int_{}^{} {3{x^2}{e^{2{x^3}}}dx} = \frac{1}{2}\int_{}^{} {{e^{2{x^3}}}\,\left( {6{x^2}} \right)dx} \hfill \\
\frac{1}{2}\int_{}^{} {{e^u}du} \hfill \\
Integrating \hfill \\
\frac{1}{2}{e^u} + C \hfill \\
Substituting\,\,u = 2{x^3}\,\,for\,\,u\,\,gives \hfill \\
\frac{1}{2}{e^{2{x^3}}} + C \hfill \\
\end{gathered} \]
