Answer
$$-\ln \dfrac{(1-x)}{x}; -1 \leq x \leq 1$$
Work Step by Step
The Taylor series for $\ln (1+x)$ can be defined as: $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$
We have
$\\=1+\dfrac{x}{2}+ \dfrac{x^2}{3} + \dfrac{x^3}{4}+....\\=(- \dfrac{1}{x})(-x- \dfrac{x^2}{2}- \dfrac{x^3}{3} - \dfrac{x^4}{4}-......)\\= (-1/x) \ln (1-x) \\=-\ln \dfrac{(1-x)}{x}; -1 \leq x \leq 1$
