6.42 Let $F=\lg \frac{1+x}{1-x}$, and replace every $x$ in the expression with $\frac{3 x+x^{3}}{1+3 x^{2}}$ to obtain a new function $G$. This $G$ simplifies to
(A) $-F$.
(B) $F$.
(C) $3 F$.
(D) $F^{2}$.
(E) $F^{3}-F$.
(14th American High School Mathematics Examination, 1963)
[Solution] From $\frac{1+\frac{3 x+x^{3}}{1+3 x^{2}}}{1-\frac{3 x+x^{3}}{1+3 x^{2}}}=\frac{1+3 x+3 x^{2}+x^{3}}{1-3 x+3 x^{2}-x^{3}}$

then
$$
=\frac{(1+x)^{3}}{(1-x)^{3}}
$$

Therefore, the answer is $(C)$.
$$
G=\operatorname{Ig}\left(\frac{1+x}{1-x}\right)^{3}=3 \lg \frac{1+x}{1-x}=3 F .
$$