13. Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3}-17 x-18=0$, with $-4<x_{1}<-3$, and $4<x_{3}<5$.
(1) Find the integer part of $x_{2}$;
(2) Find the value of $\arctan x_{1}+\arctan x_{2}+\arctan x_{3}$.
Since $x_{1}, x_{2}, x_{3}$ are the roots of the equation, we have
$$
x^{3}-17 x-18=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)
$$

By comparing the coefficients on both sides, we get
$$
x_{1}+x_{2}+x_{3}=0, x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=-17, x_{1} x_{2} x_{3}=18
$$
$\qquad$
(1) From $x_{1} \in(-4,-3)$ and $x_{3} \in(4,5)$, we know $x_{2}=-x_{1}-x_{3} \in(-2,0)$.

Note that $f(x)=x^{3}-17 x-18$ satisfies $f(0)=-180$, so $f(x)$ has a root in the interval $(-2,-1)$, i.e., $x_{2} \in(-2,-1)$. Therefore, the integer part of $x_{2}$ is -2;
(2) Let $\arctan x_{i}=\theta_{i}, i=1,2,3$, from (1) we know $\theta_{1}, \theta_{2} \in\left(-\frac{\pi}{2},-\frac{\pi}{4}\right)$, and $\theta_{3} \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, thus
$$
\theta_{1}+\theta_{2}+\theta_{3} \in\left(-\frac{3 \pi}{4}, 0\right)
$$

Note that $\tan \left(\theta_{1}+\theta_{2}\right)=\frac{\tan \theta_{1}+\tan \theta_{2}}{1-\tan \theta_{1} \tan \theta_{2}}=\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}$, hence
$$
\begin{aligned}
\tan \left(\theta_{1}+\theta_{2}+\theta_{3}\right) & =\frac{\tan \left(\theta_{1}+\theta_{2}\right)+\tan \theta_{3}}{1-\tan \left(\theta_{1}+\theta_{2}\right) \tan \theta_{3}}=\frac{\frac{\left(x_{1}+x_{2}\right)}{\left(1-x_{1} x_{2}\right)}+x_{3}}{1-\frac{x_{3}\left(x_{1}+x_{2}\right)}{1-x_{1} x_{2}}} \\
& =\frac{x_{1}+x_{2}+x_{3}-x_{1} x_{2} x_{3}}{1-\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)}=\frac{0-18}{1-(-17)}=-1
\end{aligned}
$$

This indicates that $\theta_{1}+\theta_{2}+\theta_{3}=-\frac{\pi}{4}$, i.e., $\arctan x_{1}+\arctan x_{2}+\arctan x_{3}=-\frac{\pi}{4} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ (20 points)