If $x=2+\sqrt{3}$, the value of $x+\frac{1}{x}$ is

Given $x=2+\sqrt{3}$. Let's find the value of $\frac{1}{x}$ by rationalizing the denominator:
$$\frac{1}{x}=\frac{1}{2+\sqrt{3}}$$

Multiply the numerator and denominator by the conjugate $(2-\sqrt{3})$:

$$\frac{1}{x}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=\frac{2-\sqrt{3}}{2^2-(\sqrt{3}{)}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$$

Now, compute $x+\frac{1}{x}$:

$$x+\frac{1}{x}=(2+\sqrt{3})+(2-\sqrt{3})=4$$