How to show that $\ln{x}=2\sum_{n=0}^{\infty}\frac{1}{(2n+1)}\left(\frac{x-1}{x+1}\right)^{2n+1}$? 
How do I show that $\displaystyle\ln{x}=2\sum_{n=0}^{\infty}\frac{1}{(2n+1)}\left(\frac{x-1}{x+1}\right)^{2n+1},\ x>0$?

I have tried using Mclaurin series of $\ln(1+x)$ and replacing $x$ with $\dfrac{x-1}{x+1}$ in it but that didn't worked.

Simplify the expression inside the sum using an integral of a much more elementary function in the following way:
\begin{align*}
2\sum_{n=0}^{\infty}\frac 1{2n+1}\left(\frac{x-1}{x+1}\right)^{2n+1}&=
2\sum_{n=0}^{\infty}\int_0^{\frac{x-1}{x+1}}y^{2n}\,\mathrm dy=
2\int_0^{\frac{x-1}{x+1}}\sum_{n=0}^{\infty}y^{2n}\,\mathrm dy=
2\int_0^{\frac{x-1}{x+1}}\frac{\mathrm dy}{1-y^2}\\[10pt]
&=\left.\ln\left(\frac{1+y}{1-y}\right)\right|_{y=0}^{y=\frac{x-1}{x+1}}=\ln x
\end{align*}
