2. The sum of the coefficients of the terms with integer powers of $x$ in the expansion of $(\sqrt{x}+2)^{2 n+1}$ is

保留源文本的换行和格式，直接输出翻译结果。
$$
\frac{1}{2}\left(3^{2 n+1}+1\right)
$$
2. 【Analysis and Solution】Let $(\sqrt{x}+2)^{2 n+1}=f(x)+\sqrt{x} g(x)$, where $f(x)$ and $g(x)$ are polynomials of $x$, then what we need to find is $f(1)$. And $(2+\sqrt{x})^{2 n+1}+(2-\sqrt{x})^{2 n+1}=f(x)+\sqrt{x} g(x)+f(x)-\sqrt{x} g(x)$.
$$
\begin{aligned}
& \left.\therefore f(1)=\frac{1}{2}\left[(2+\sqrt{1})^{2 n+1}+(2-\sqrt{1})^{2 n+1}\right)\right] \\
= & \frac{1}{2}\left(3^{2 n+1}+1\right) .
\end{aligned}
$$