Answer
\[\frac{1}{2}\,{\left( {2{y^3} + 3{y^2} + 1} \right)^{\frac{1}{3}}}\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{y^2} + y}}{{\,{{\left( {2{y^3} + 3{y^2} + 1} \right)}^{2/3}}}}dy} \hfill \\
Let\,\,\,u = 2{y^3} + 3{y^2} + 1\,\,,\,\,So\,\,that \hfill \\
du = \,\left( {6{y^2} + 6y} \right)dy \hfill \\
du = 6\,\left( {{y^2} + 6} \right)dy \hfill \\
Then \hfill \\
\frac{1}{6}\int_{}^{} {\frac{{6\,\left( {{y^2} + y} \right)dy}}{{\,{{\left( {2{y^3} + 3{y^2} + 1} \right)}^{2/3}}}} = \frac{1}{6}\int_{}^{} {\frac{{du}}{{{u^{2/3}}}}} } \hfill \\
\frac{1}{6}\int_{}^{} {{u^{ - 2/3}}du} \hfill \\
Power\,\,rule \hfill \\
\frac{1}{6}\,\left( {\frac{{{u^{1/2}}}}{{1/3}}} \right) + C \hfill \\
\frac{1}{2}{u^{\frac{1}{3}}} + C \hfill \\
Substituting\,\,u = 2{y^3} + 3{y^2} + 1\,\,gives \hfill \\
\frac{1}{2}\,{\left( {2{y^3} + 3{y^2} + 1} \right)^{\frac{1}{3}}} \hfill \\
\end{gathered} \]
