Answer
\[\frac{2}{3}{y^{\frac{1}{2}}} - \frac{1}{4}{y^2} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{2{y^{\frac{1}{2}}} - 3{y^2}}}{{6y}}dy} \hfill \\
rewrite\,\,the\,\,integrand\,\,as\,\,follows \hfill \\
\int_{}^{} \begin{gathered}
\frac{{2{y^{\frac{1}{2}}}}}{{6y}}dy - \int_{}^{} {\frac{{3{y^2}}}{{6y}}dy} \hfill \\
Simplifying \hfill \\
\end{gathered} \hfill \\
\frac{1}{3}\int_{}^{} {{y^{ - \frac{1}{2}}}dy - \frac{1}{2}\int_{}^{} {ydy} } \hfill \\
Use\,\,power\,\,rule\,\,\int_{}^{} {{y^n}dy} = \frac{{{y^{n + 1}}}}{{n + 1}} + C \hfill \\
\frac{1}{3}\,\left( {\frac{{{1^{\frac{1}{2}}}}}{{1/2}}} \right) - \frac{1}{2}\,\left( {\frac{{{y^2}}}{2}} \right) + C \hfill \\
Then \hfill \\
\frac{2}{3}{y^{\frac{1}{2}}} - \frac{1}{4}{y^2} + C \hfill \\
\end{gathered} \]
