smc

Replace the $x$ in $F$ with $\frac{3x+x^3}{1+3x^2}$. $$\log \frac{1+\frac{3x+x^3}{1+3x^2}}{1-\frac{3x+x^3}{1+3x^2}}$$ $$\log \frac{\frac{1+3x^2}{1+3x^2}+\frac{3x+x^3}{1+3x^2}}{\frac{1+3x^2}{1+3x^2}-\frac{3x+x^3}{1+3x^2}}$$ $$\log (\frac{1+3x+3x^2+x^3}{1+3x^2}÷\frac{1-3x+3x^2-x^3}{1+3x^2})$$ Recall that from the Binomial Theorem, $(1+x{)}^3=1+3x+3x^2+x^3$ and $(1-x{)}^3=1-3x+3x^2-x^3$. $$\log (\frac{(1+x{)}^3}{1+3x^2}÷\frac{(1-x{)}^3}{1+3x^2})$$ $$\log \frac{(1+x{)}^3}{(1-x{)}^3}$$ $$3\log \frac{1+x}{1-x}$$ Thus, the expression is equivalent to $3F$, which is answer choice $\boxed{3F}$.