6. The sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}$. Let $[x]$ denote the greatest integer not exceeding $x$, then $\left[\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{2009}}\right]$ is $\qquad$ .
6. 1 Detailed Explanation: From the problem, we know that $x_{n+1}>x_{n}, x_{2}=\frac{3}{4}, x_{3}=\frac{21}{16}>1$, so $x_{2010}>1, 0<\frac{1}{x_{2010}}<1$,
$$
\begin{array}{l}
\text { Also, } \frac{1}{x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{1+x_{n}} \Rightarrow \frac{1}{1+x_{n}}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}}, \\
\text { Therefore, }\left[\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{2009}}\right]=\left[\frac{1}{x_{1}}-\frac{1}{x_{2}}+\frac{1}{x_{2}}-\frac{1}{x_{3}}+\cdots+\frac{1}{x_{2009}}-\frac{1}{x_{2010}}\right]=\left[2-\frac{1}{x_{2010}}\right]=1 .
\end{array}
$$