Answer
$\ln(\frac{1+x}{1-x})=2x+\frac{2x^3}{3}+\frac{2x^5}{5}+...\frac{2x^{2n+1}}{2n+1}+...$
Work Step by Step
The Taylor series for $\ln(1+x)$ is
$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...+\frac{(-1)^nx^{n+1}}{n+1}...$
So, the Taylor series for $f(x)=ln(1-x)=\ln[1+(-x)]$ is
$-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+...+\frac{x^{n+1}}{n+1}...$
Hence
$\ln(1+x)-\ln(1-x)$
$=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...+\frac{(-1)^nx^{n+1}}{n+1}...-[-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+...+\frac{x^{n+1}}{n+1}...]$
$=2x+\frac{2x^3}{3}+\frac{2x^5}{5}+...\frac{2x^{2n+1}}{2n+1}+...$
We are given $\ln(\frac{1+x}{1-x})=\ln(1+x)-\ln(1-x)$
so $\ln(\frac{1+x}{1-x})=2x+\frac{2x^3}{3}+\frac{2x^5}{5}+...\frac{2x^{2n+1}}{2n+1}+...$
