One. (20 points) Given $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$. Find the integer part of $x^{6}+x^{5}+2 x^{4}-4 x^{3}+3 x^{2}+4 x-4$.


Given $x>0$.
If $\sqrt{1+x}>x$, then
$$
\begin{array}{l}
x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}>\sqrt{1+\sqrt{1+x}} \\
>\sqrt{1+x},
\end{array}
$$

which contradicts the assumption;
If $\sqrt{1+x}<x$, then
$$
\begin{array}{l}
x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}<\sqrt{1+\sqrt{1+x}} \\
<\sqrt{1+x},
\end{array}
$$

which also contradicts the assumption.
Therefore, $\sqrt{1+x}=x$.
Solving gives $x=\frac{1+\sqrt{5}}{2}$ (negative value is discarded).
$$
\begin{aligned}
\text { Hence the original expression }= & \left(x^{2}-x-1\right)\left(x^{4}+2 x^{3}+5 x^{2}+\right. \\
& 3 x+11)+18 x+7 \\
= & 18 \times \frac{1+\sqrt{5}}{2}+7=16+9 \sqrt{5}=16+\sqrt{405} .
\end{aligned}
$$

Since $20<\sqrt{405}<21$, the integer part of the original expression is 36.