Answer
$$2\sqrt{x}e^{\sqrt{x}}-2e^{\sqrt{x}}+c$$
Work Step by Step
Given $$\int e^{\sqrt{x}}dx$$
Let $s^2=x\ \to \ 2sds=dx$, then
$$ \int e^{\sqrt{x}}dx = 2\int se^sds$$
Let
\begin{align*}
u&=s\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=e^sds\\
u&= ds\ \ \ \ \ \ \ \ \ \ \ \ \ dv= e^s
\end{align*}
Then using integration by parts
\begin{align*}
2\int se^sds &=2\left(uv-\int vdu\right)\\
&=2\left(se^s-\int e^sds\right)\\
&=2se^s-2e^s +c\\
&=2\sqrt{x}e^{\sqrt{x}}-2e^{\sqrt{x}}+c
\end{align*}
