$7 \cdot 115$ If $f(x)=\lg \left(\frac{1+x}{1-x}\right),-1<x<1$, then the expression $f\left(\frac{3 x+x^{3}}{1+3 x^{2}}\right)$ in terms of $f(x)$ is
(A) $-f(x)$.
(B) $2 f(x)$.
(C) $3 f(x)$.
(D) $[f(x)]^{2}$.
(E) $[f(x)]^{3}-f(x)$.
(23rd American High School Mathematics Examination, 1972)
[Solution]
$$
\begin{aligned}
f\left(\frac{3 x+x^{3}}{1+3 x^{2}}\right) & =\lg \frac{1+\frac{3 x+x^{3}}{1+3 x^{2}}}{1-\frac{3 x+x^{3}}{1+3 x^{2}}}=\lg \frac{1+3 x^{2}+3 x+x^{3}}{1+3 x^{2}-3 x-x^{3}} \\
& =\lg \frac{(1+x)^{3}}{(1-x)^{3}}=\lg \left(\frac{1+x}{1-x}\right)^{3}=3 \lg \frac{1+x}{1-x} \\
& =3 f(x)
\end{aligned}
$$

Therefore, the correct choice is $(C)$.