Answer
\[ = \frac{{\,{{\left( {3{y^3} + 1} \right)}^5}}}{{45}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{y^2}\,{{\left( {3{y^3} + 1} \right)}^4}dy} \hfill \\
\hfill \\
set\,\,u = 3{y^3} + 1 \hfill \\
du = 9{y^2}dy \hfill \\
\frac{1}{9}du = {y^2}dy \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
\int_{}^{} {{y^2}\,{{\left( {3{y^3} + 1} \right)}^4}dy} = \frac{1}{9}\int {{u^4}du} \hfill \\
\hfill \\
integrate \hfill \\
\frac{1}{9}{u^5} + C \hfill \\
\hfill \\
Therefore, \hfill \\
\hfill \\
= \frac{1}{9}\,\left( {\frac{{\,{{\left( {3{y^3} + 1} \right)}^5}}}{5}} \right) + C \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{{\,{{\left( {3{y^3} + 1} \right)}^5}}}{{45}} + C \hfill \\
\end{gathered} \]