(9) Let the function $f(x)=\frac{1}{2}+\log _{2} \frac{x}{1-x}, S_{n}=\sum_{i=1}^{n-1} f\left(\frac{i}{n}\right)$, where $n \in$ $\mathbf{N}^{*}$, and $n \geqslant 2$, then $S_{n}=$ $\qquad$ .
(9) $\frac{n-1}{2}$ Hint: When $x_{1}+x_{2}=1$,
$$
\begin{array}{c}
f\left(x_{1}\right)+f\left(x_{2}\right)=1+\log _{2} \frac{x_{1} x_{2}}{\left(1-x_{1}\right)\left(1-x_{2}\right)}=1, \\
2 S_{n}=\sum_{i=1}^{n-1}\left[f\left(\frac{i}{n}\right)+f\left(\frac{n-i}{n}\right)\right]=n-1,
\end{array}
$$

Therefore, $S_{n}=\frac{n-1}{2}$.