Answer
$(2) \Sigma_{n=0}^\infty \dfrac{x^{2n+1}}{2n+1}$
Work Step by Step
We know that the Maclaurin Series for $ \ln (1+x)$ is defined as:
$ \ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...$ and $ \ln (1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-...$
Now, $ \ln (1+x)-\ln (1-x)=(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...)-(-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-...)$
Thus, $=2x+\dfrac{2x^3}{3}+\dfrac{2x^5}{5}+...$
or, $(2) \Sigma_{n=0}^\infty \dfrac{x^{2n+1}}{2n+1}$
