5. The sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{k+1}=x_{k}^{2}+x_{k}$, then the integer part of $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{2009}+1}$ is . $\qquad$
5.1 Detailed Explanation: From the given, we know that $\left\{x_{n}\right\}$ is an increasing sequence, and from $x_{k+1}=x_{k}^{2}+x_{k}$, we have $\frac{1}{x_{k+1}}=\frac{1}{x_{k}^{2}+x_{k}}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, which gives $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$. Therefore, $S=\sum_{k=1}^{2009} \frac{1}{x_{k}+1}=\sum_{k=1}^{2009}\left(\frac{1}{x_{k}}-\frac{1}{x_{k+1}}\right)=\frac{1}{x_{1}}-\frac{1}{x_{2010}}=2-\frac{1}{x_{2010}}$. Since $x_{2010}>x_{3}>1$, we have $1<S<2$, hence $[S]=1$.