Answer
$$ 2xe^{\sqrt{x}} -4[\sqrt{x}e^{\sqrt{x}}-e^{\sqrt{x}}]+C$$
Work Step by Step
Given $$\int \sqrt{x}e^{\sqrt{x}} dx$$
Let $$z^2 = x\ \ \ \ \to\ \ \ 2zdz= dx$$
Then
$$ \int 2z^2e^{z} dz $$
Let
\begin{align*}
u&=2z^2 \ \ \ \ \ \ \ dv=e^zdz\\
du&=4zdz \ \ \ \ \ \ v= e^z
\end{align*}
Then
\begin{align*}
\int \sqrt{x}e^{\sqrt{x}} dx&=\int 2z^2e^{z} dz\\
&= 2z^2 e^z -4\int ze^zdz \\
\text{integrate by parts again }\\
&= 2z^2 e^z -4[ze^z-e^z]+C\\
&= 2xe^{\sqrt{x}} -4[\sqrt{x}e^{\sqrt{x}}-e^{\sqrt{x}}]+C
\end{align*}
