5. A. Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{99^{3}}$. Then the integer part of $4 S$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7
5. A. A.

When $k=2,3, \cdots, 99$, we have
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {, }
$$

Thus, $1<S=1+\sum_{k=2}^{29} \frac{1}{k^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99 \times 100}\right)<\frac{5}{4} \text {. }
$$

Therefore, $4<4 S<5$.
Hence, the integer part of $4 S$ is 4.