5. If $\sqrt{3}$ is between $\frac{x+3}{x}$ and $\frac{x+4}{x+1}$, then the positive integer $x$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
5. B.

Solution: From the given, we have $\frac{x+3}{x}<\sqrt{3}<\frac{x+4}{x+1}$ or $\frac{x+4}{x+1}<\sqrt{3}<\frac{x+3}{x}$.
(1) No solution. From (2), we solve to get
$$
\frac{4-\sqrt{3}}{\sqrt{3}-1}<x<\frac{3}{\sqrt{3}-1}. [x]=4.
$$