Define the sequence $(x_{n})$: $x_{1}=\frac{1}{3}$ and $x_{n+1}=x_{n}^{2}+x_{n}$. Find $\left[\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\dots+\frac{1}{x_{2007}+1}\right]$, wehere $[$ $]$ denotes the integer part.
1. We start with the given sequence \( (x_n) \) defined by \( x_1 = \frac{1}{3} \) and \( x_{n+1} = x_n^2 + x_n \). We can rewrite the recurrence relation as:
   \[
   x_{n+1} = x_n (x_n + 1)
   \]

2. We need to find the sum \( \left\lfloor \sum_{k=1}^{2007} \frac{1}{x_k + 1} \right\rfloor \), where \( \lfloor \cdot \rfloor \) denotes the integer part.

3. Notice that:
   \[
   \frac{1}{x_{n+1}} = \frac{1}{x_n (x_n + 1)} = \frac{1}{x_n} - \frac{1}{x_n + 1}
   \]
   Therefore:
   \[
   \frac{1}{x_n + 1} = \frac{1}{x_n} - \frac{1}{x_{n+1}}
   \]

4. Summing both sides from \( k = 1 \) to \( k = 2007 \), we get:
   \[
   \sum_{k=1}^{2007} \frac{1}{x_k + 1} = \sum_{k=1}^{2007} \left( \frac{1}{x_k} - \frac{1}{x_{k+1}} \right)
   \]

5. This is a telescoping series, so most terms cancel out, leaving:
   \[
   \sum_{k=1}^{2007} \frac{1}{x_k + 1} = \frac{1}{x_1} - \frac{1}{x_{2008}}
   \]

6. Given \( x_1 = \frac{1}{3} \), we have:
   \[
   \frac{1}{x_1} = 3
   \]

7. To estimate \( x_{2008} \), observe that \( x_n \) grows very quickly. For example:
   \[
   x_2 = x_1 (x_1 + 1) = \frac{1}{3} \left( \frac{1}{3} + 1 \right) = \frac{1}{3} \cdot \frac{4}{3} = \frac{4}{9}
   \]
   \[
   x_3 = x_2 (x_2 + 1) = \frac{4}{9} \left( \frac{4}{9} + 1 \right) = \frac{4}{9} \cdot \frac{13}{9} = \frac{52}{81}
   \]
   \[
   x_4 = x_3 (x_3 + 1) = \frac{52}{81} \left( \frac{52}{81} + 1 \right) = \frac{52}{81} \cdot \frac{133}{81} = \frac{6916}{6561} > 1
   \]

8. Since \( x_n \) grows exponentially, \( x_{2008} \) will be extremely large. Specifically, \( x_{2008} > 2^{2004} \).

9. Therefore, \( \frac{1}{x_{2008}} \) will be very small, approaching 0. Thus:
   \[
   \frac{1}{x_1} - \frac{1}{x_{2008}} \approx 3 - 0 = 3
   \]

10. Taking the integer part, we get:
   \[
   \left\lfloor 3 - \frac{1}{x_{2008}} \right\rfloor = 2
   \]

The final answer is \(\boxed{2}\)