Answer
$$2 e^{\sqrt{x}}(\sqrt{x}-1)+C $$
Work Step by Step
Given $$\int e^{\sqrt{x}} d x $$
Let $u=x^{1/2}\ \ \ \ \ du = \frac{1}{2\sqrt{x}}dx$, then
\begin{aligned}
\int e^{\sqrt{x}} d x &=\int e^{u} \cdot 2 u d u \\ &=2 \int u e^{u} d u \end{aligned}
Let \begin{align*}
u&=u \ \ \ \ \ \ \ dv=e^udu\\
du&=du\ \ \ \ \ \ v =e^u
\end{align*}
Then
\begin{aligned} 2 \int u e^{u} d u &=2\left[u \cdot e^{u}-\int 1 \cdot e^{u} d u\right] \\ &=2 u e^{u}-2 \int e^{u} d u \\ &=2 u e^{u}-2 e^{u}+C \\ &=2 \sqrt{x} e^{\sqrt{x}}-2 e^{\sqrt{x}}+C \\ &=2 e^{\sqrt{x}}(\sqrt{x}-1)+C \end{aligned}
