Find consecutive integers bounding the expression \[\frac{1}{x_1 + 1}+\frac{1}{x_2 + 1}+\frac{1}{x_3 + 1}+... +\frac{1}{x_{2001} + 1}+\frac{1}{x_{2002} + 1}\]
where $x_1 = 1/3$ and $x_{n+1} = x_n^2 + x_n.$
1. Given the sequence \( x_1 = \frac{1}{3} \) and \( x_{n+1} = x_n^2 + x_n \), we need to find consecutive integers bounding the sum:
   \[
   S = \sum_{k=1}^{2002} \frac{1}{x_k + 1}
   \]

2. Using the hint, we can rewrite \( x_{k+1} \) as:
   \[
   x_{k+1} = x_k^2 + x_k = x_k(1 + x_k)
   \]
   This implies:
   \[
   \frac{1}{1 + x_k} = \frac{1}{x_k} - \frac{1}{x_{k+1}}
   \]

3. Therefore, the sum \( S \) can be expressed as a telescoping series:
   \[
   S = \sum_{k=1}^{2002} \left( \frac{1}{x_k} - \frac{1}{x_{k+1}} \right)
   \]

4. Notice that in a telescoping series, most terms cancel out:
   \[
   S = \left( \frac{1}{x_1} - \frac{1}{x_2} \right) + \left( \frac{1}{x_2} - \frac{1}{x_3} \right) + \cdots + \left( \frac{1}{x_{2001}} - \frac{1}{x_{2002}} \right) + \left( \frac{1}{x_{2002}} - \frac{1}{x_{2003}} \right)
   \]
   This simplifies to:
   \[
   S = \frac{1}{x_1} - \frac{1}{x_{2003}}
   \]

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

6. To find \( \frac{1}{x_{2003}} \), we observe that \( x_n \) grows very quickly. For large \( n \), \( x_n \) becomes very large, making \( \frac{1}{x_{2003}} \) very small. Thus, \( \frac{1}{x_{2003}} \approx 0 \).

7. Therefore, the sum \( S \) is approximately:
   \[
   S \approx 3
   \]

8. Since \( \frac{1}{x_{2003}} \) is positive but very small, \( S \) is slightly less than 3. Hence, the consecutive integers bounding \( S \) are 2 and 3.

The final answer is \( \boxed{2 \text{ and } 3} \)