Answer
\[2[x-2\sqrt x+2]e^{\sqrt x}\]
Work Step by Step
Let \[I=\int\sqrt x\:e^{\sqrt x}dx\;\;\;\ldots(1)\]
Substitute $\sqrt x=t\;\;\;\ldots (2)$
\[\Rightarrow \frac{1}{2\sqrt x}dx=dt\]
\[I=\int 2t^2e^t\:dt\]
Using integration by parts
\[I=2\left[t^2\int e^t dt-\int \left((t^2)'\int e^t dt\right)dt\right]+C\]
Where $C$ is constant of integration
\[I=2\left[t^2 e^t-2\int te^t dt\right]\;\;\;\ldots (3)\]
Let \[I_1=\int te^tdt\]
\[I_1=t\int e^tdt-\int \left((t)'\int e^tdt\right)dt\]
\[I_1=te^t-\int e^tdt\]
\[I_1=te^t-e^t\;\;\;\ldots (4)\]
Using (4) in (3)
\[I=2\left[t^2 e^t-2 te^t+2e^t\right]\]
\[I=2[t^2-2t+2]e^t\]
From (2)
\[I=2[x-2\sqrt x+2]e^{\sqrt x}\]
Hence, \[\;I=2[x-2\sqrt x+2]e^{\sqrt x}.\]
