8. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{3}$, and for any $n \in \mathbf{N}^{*}, a_{n+1}=a_{n}^{2}+a_{n}$, then the integer part of $\sum_{n=1}^{2016} \frac{1}{a_{n}+1}$ is
$$
a_{n+1}=a_{n}\left(a_{n}+1\right) \Rightarrow \frac{1}{a_{n+1}}=\frac{1}{a_{n}}-\frac{1}{a_{n}+1} \Rightarrow \frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \text {, }
$$

Thus, $\sum_{n=1}^{2016} \frac{1}{a_{n}+1}=\frac{1}{a_{1}}-\frac{1}{a_{2017}}=3-\frac{1}{a_{2017}}$. Also, $a_{n+1}-a_{n}=a_{n}^{2}>0$, and $a_{2}=\frac{1}{3} \cdot \frac{4}{3}=\frac{4}{9}, a_{3}=\frac{4}{9} \cdot \frac{13}{9}=\frac{52}{81}, a_{4}=\frac{52}{81} \cdot \frac{133}{81}=\frac{6916}{6561}>1$, hence $a_{2017}>1 \Rightarrow 0<\frac{1}{a_{2017}}<1$, so the integer part of $\sum_{n=1}^{2016} \frac{1}{a_{n}+1}$ is 2.