Answer
\[{e^{\sqrt y }} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{e^{\sqrt y }}}}{{2\sqrt y }}dy} \hfill \\
Let\,\,u = \sqrt y = {y^{\frac{1}{2}}} \hfill \\
So\,\,that \hfill \\
du = \frac{1}{2}{y^{ - 1/2}} = \frac{1}{{2\sqrt y }} \hfill \\
Then\,\,\int_{}^{} {\frac{{{e^{\sqrt y }}}}{{2\sqrt y }} = \int_{}^{} {{e^u}du} } \hfill \\
Integrating\, \hfill \\
{e^u} + C \hfill \\
Substituting\,\,u = \sqrt y \,\,for\,\,u\,\,gives\,\, \hfill \\
{e^{\sqrt y }} + C \hfill \\
\hfill \\
\end{gathered} \]
