Answer
$$\int e^{\sqrt{2 x}} d x =(\sqrt{2x}-1)e^{\sqrt{2x}} +c$$
Work Step by Step
$$
\int e^{\sqrt{2 x}} d x
$$
Let $ z^2= 2x\ \Rightarrow \ \ zdz= dx $, then
$$
\int e^{\sqrt{2 x}} d x = \int ze^{z} d z
$$
Integrate by parts
\begin{align*}
u&=z\ \ \ \ \ \ dv=e^zdz\\
du&=dz\ \ \ \ \ \ v=e^z
\end{align*}
Then
\begin{align*}
\int e^{\sqrt{2 x}} d x& = \int ze^{z} d z\\
&=ze^z-e^z+c\\
&=\sqrt{2x}e^{\sqrt{2x}}-e^{\sqrt{2x}}+c\\
&=(\sqrt{2x}-1)e^{\sqrt{2x}} +c
\end{align*}
