5. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1$, and subsequent terms are given by the formula $a_{n}=a_{n-1}+\frac{1}{n(n-1)}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right)$, then $a_{4}=$
A. $\frac{7}{4}$
B. $-\frac{7}{4}$
C. $\frac{4}{7}$
D. $-\frac{4}{7}$
$$
a_{n}=a_{n-1}+\frac{1}{n-1}-\frac{1}{n} \Rightarrow a_{n}+\frac{1}{n}=a_{n-1}+\frac{1}{n-1}=\cdots=a_{1}+1=2 \Rightarrow a_{n}=2-\frac{1}{n} \text {, }
$$

Thus $a_{4}=2-\frac{1}{4}=\frac{7}{4}$. Therefore, the answer is $A$.