We all know the series expansion $$\log 2=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n. \tag1$$ I also am able to use the method of Wilf-Zeilberger to the effect that $$\log 2=3\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n\binom{2n}n2^n}. \tag2$$

QUESTION. Can you provide yet another proof of the formula in (2)?

Remark. My motivation for this question goes beyond this particular series, hoping it paves a way forward in my study.

Postscript. After those generous replies (see below), it appears that the idea rests on $$\log\left(1+\frac1x\right)=2\sinh^{-1}\left(\frac1{2\sqrt{x+x^2}}\right)$$ so that we may put $x=1$ to obtain (1) and (2).