Answer
\[ = {e^2}^{\sqrt y + 1} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{e^{2\sqrt y + 1}}}}{{\sqrt y }}dy} \hfill \\
\hfill \\
set\,\,\,\left( {2\sqrt y + 1} \right) = u\,\,then\,\,\frac{{dy}}{{\sqrt y }} = du \hfill \\
\hfill \\
integral\,\,becomes \hfill \\
\hfill \\
\int_{}^{} {\frac{{{e^{2\sqrt y + 1}}}}{{\sqrt y }}dy} = \int_{}^{} {{e^u}du} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= {e^u} + C \hfill \\
\hfill \\
substituting\,back\,\,\,u = 2\sqrt y + 1 \hfill \\
\hfill \\
= {e^2}^{\sqrt y + 1} + C \hfill \\
\end{gathered} \]
