Using the Maclaurin expansion for $\ln(1+x)$

Question

I don't know how to use the Maclaurin expansion for $\ln(1+x)$ to prove that : $$\ln\left(\frac{1+x}{1-x}\right)= 2x\left(1+\frac{x^2}{3}+\frac{x^4}{5}+\frac{x^6}{7}+\dots\right).$$ I know that: $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\dots$$

Answer 1

Hint:

Assuming $\;|x|<1\;$ :

$$\log\frac{1+x}{1-x}=\log(1+x)-\log(1-x)=\sum_{n=1}^\infty\left((-1)^n\frac{x^n}n-(-1)^{n+1}\frac{x^n}n\right)=\ldots $$