If $f(x)=\log \left(\cfrac{1+x}{1-x}\right)$ for $-1 < x < 1$,then find $f \left(\cfrac{3x+x^3}{1+3x^2}\right)$ in terms of $f(x)$.

Question

If $f(x)=\log \left(\cfrac{1+x}{1-x}\right)$ for $-1 < x < 1$,then find $f \left(\cfrac{3x+x^3}{1+3x^2}\right)$ in terms of $f(x)$.

My Attempt $$f \left(\cfrac{3x+x^3}{1+3x^2}\right)=\log\left(\cfrac{1+\cfrac{3x+x^3}{1+3x^2}}{1-\cfrac{3x+x^3}{1+3x^2}}\right)=\log \left(\cfrac{1+3x^2+3x+x^3}{1+3x^2-3x-x^3}\right)=\\\log(1+3x^2+3x+x^3)-\log(1+3x^2-3x-x^3)$$

Now I am kinda clueless about how to express this in terms of $f(x)$.

Can you guys help ?

Answer 1

$$\log(1+3x^2+3x+x^3)-\log(1+3x^2-3x-x^3)=\log(1+x)^3-\log(1-x)^3=3[\log(1+x)-\log(1-x)]=3f(x)$$

Answer 2

hint $$log(\frac{1+x}{1-x})=2(x+\frac{x^3}{3!}...\infty)$$ for $|x|<1$