I have to prove for $|x| < 1$ that $$ \ln\frac{2(1-\sqrt{1-x})}{x} = \frac 12 \cdot \frac x2 + \frac 12 \cdot \frac 34 \cdot \frac{x^2}{4} + \frac 12 \cdot \frac 34 \cdot \frac 56 \cdot \frac{x^3}{6} + \dots $$
I tried to use the Taylor series for the left side, but derivatives become too complicated. How can I prove this equality?
The function $u(x)$ to be expanded is such that $u(0)=0$ and $$u'(x)=\frac1{2x}\left(\frac1{\sqrt{1-x}}-1\right).$$ Assuming one knows that $$\frac1{\sqrt{1-x}}=\sum_{n\geqslant0}a_nx^n,$$ with $a_0=1$, this yields $$u(x)=\sum_{n\geqslant0}\frac{a_n}{2n}x^n.$$ Can you identify $(a_n)_{n\geqslant1}$? Hint:$$\frac1{\sqrt{1-x}}=(1-x)^\alpha,\qquad \alpha=-\frac12.$$