Sequence {$x_n$} satisfies: $x_1=1$ , ${x_n=\sqrt{x_{n-1}^2+x_{n-1}}+x_{n-1}}$ ( ${n>=2}$ )
Find the general term of {$x_n$}
1. Given the sequence $\{x_n\}$ defined by:
   \[
   x_1 = 1
   \]
   \[
   x_n = \sqrt{x_{n-1}^2 + x_{n-1}} + x_{n-1} \quad \text{for} \quad n \geq 2
   \]

2. Squaring both sides of the recurrence relation:
   \[
   x_n^2 = (\sqrt{x_{n-1}^2 + x_{n-1}} + x_{n-1})^2
   \]
   Expanding the right-hand side:
   \[
   x_n^2 = x_{n-1}^2 + x_{n-1} + 2x_{n-1}\sqrt{x_{n-1}^2 + x_{n-1}} + x_{n-1}^2
   \]
   Simplifying:
   \[
   x_n^2 = 2x_{n-1}^2 + x_{n-1} + 2x_{n-1}\sqrt{x_{n-1}^2 + x_{n-1}}
   \]

3. Rearranging the terms:
   \[
   x_n^2 - 2x_{n-1}^2 - x_{n-1} = 2x_{n-1}\sqrt{x_{n-1}^2 + x_{n-1}}
   \]
   Dividing both sides by $2x_{n-1}$:
   \[
   \frac{x_n^2 - 2x_{n-1}^2 - x_{n-1}}{2x_{n-1}} = \sqrt{x_{n-1}^2 + x_{n-1}}
   \]

4. Introducing a substitution $x_n = \frac{1}{y_n}$:
   \[
   \left(\frac{1}{y_n}\right)^2 - 2\left(\frac{1}{y_{n-1}}\right)^2 - \frac{1}{y_{n-1}} = 0
   \]
   Simplifying:
   \[
   \frac{1}{y_n^2} - \frac{2}{y_{n-1}^2} - \frac{1}{y_{n-1}} = 0
   \]
   Multiplying through by $y_n^2 y_{n-1}^2$:
   \[
   y_{n-1}^2 - 2y_n^2 y_{n-1} - y_n^2 y_{n-1} = 0
   \]

5. Letting $y_n + 1 = z_n$:
   \[
   z_n^2 = z_{n-1}
   \]
   Solving this recurrence relation by induction:
   \[
   z_n = \sqrt[2^{n-1}]{2}
   \]

6. Substituting back:
   \[
   y_n = \sqrt[2^{n-1}]{2} - 1
   \]
   \[
   x_n = \frac{1}{y_n} = \frac{1}{\sqrt[2^{n-1}]{2} - 1}
   \]

The final answer is $\boxed{\frac{1}{\sqrt[2^{n-1}]{2} - 1}}$