Answer
$$2(x+\dfrac{x^3}{3}+ \dfrac{x^5}{5} +....); -1 \lt x \lt 1$$
Work Step by Step
The Taylor series for $\ln (1+x)$ can be defined as: $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$ and $\ln (1-x)=-x-\dfrac{ x^2}{2}-\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$
We have
$\ln (1+x) -\ln (1-x)=(x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....) - (-x-\dfrac{ x^2}{2}-\dfrac{x^3}{3}-....) $
or, $\ln (\dfrac{1+x}{1-x})=2(x+\dfrac{x^3}{3}+ \dfrac{x^5}{5} +....); -1 \lt x \lt 1$
