Let $F=\log\dfrac{1+x}{1-x}$. Find a new function $G$ by replacing each $x$ in $F$ by $\dfrac{3x+x^3}{1+3x^2}$, and simplify. 
The simplified expression $G$ is equal to:
$\textbf{(A)}\ -F \qquad \textbf{(B)}\ F\qquad \textbf{(C)}\ 3F \qquad \textbf{(D)}\ F^3 \qquad \textbf{(E)}\ F^3-F$
Replace the $x$ in $F$ with $\frac{3x+x^3}{1+3x^2}$.
\[\log\dfrac{1 + \frac{3x+x^3}{1+3x^2} }{1 - \frac{3x+x^3}{1+3x^2} }\]
\[\log\dfrac{\frac{1+3x^2}{1+3x^2} + \frac{3x+x^3}{1+3x^2} }{\frac{1+3x^2}{1+3x^2} - \frac{3x+x^3}{1+3x^2} }\]
\[\log (\frac{1+3x+3x^2+x^3}{1+3x^2} \div \frac{1-3x+3x^2-x^3}{1+3x^2})\]
Recall that from the [Binomial Theorem](https://artofproblemsolving.com/wiki/index.php/Binomial_Theorem), $(1+x)^3 = 1+3x+3x^2+x^3$ and $(1-x)^3 = 1-3x+3x^2-x^3$.
\[\log (\frac{(1+x)^3}{1+3x^2} \div \frac{(1-x)^3}{1+3x^2})\]
\[\log\dfrac{(1+x)^3}{(1-x)^3}\]
\[3 \log\dfrac{1+x}{1-x}\]
Thus, the expression is equivalent to $3F$, which is answer choice $\boxed{\textbf{(C)}}$.