Answer
\[2\sqrt x {e^{\sqrt x }} - 2{e^{\sqrt x }} + C\]
Work Step by Step
\[\begin{gathered}
\int {{e^{\sqrt x }}} dx \hfill \\
{\text{Let }}z = \sqrt x ,{\text{ }}dz = \frac{1}{{2\sqrt x }}dx = \frac{1}{{2z}}dx{\text{ }} \Rightarrow dx = 2zdz,{\text{ then}} \hfill \\
\int {{e^{\sqrt x }}} dx = \int {2z{e^z}} dz \hfill \\
{\text{Integrate by parts, }} \hfill \\
{\text{let }}u = 2z{\text{ }}du = 2dz \hfill \\
dv = {e^z}dz{\text{ }}v = {e^z} \hfill \\
{\text{Therefore,}} \hfill \\
\int {2z{e^z}} dz = 2z{e^z} - \int {{e^z}\left( 2 \right)} dz \hfill \\
{\text{ }} = 2z{e^z} - 2\int {{e^z}} dz \hfill \\
{\text{ }} = 2z{e^z} - 2{e^z} + C \hfill \\
{\text{Back - substitute }}z = \sqrt x \hfill \\
{\text{ }} = 2\sqrt x {e^{\sqrt x }} - 2{e^{\sqrt x }} + C \hfill \\
\end{gathered} \]
