Answer
$-\ln \dfrac{(1-x)}{x}; -1 \leq x \leq 1$
Work Step by Step
Recall the Taylor series for $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$
Now,
$=1+\dfrac{x}{2}+ \dfrac{x^2}{3} + \dfrac{x^3}{4}+....$
or, $=(- \dfrac{1}{x})(-x- \dfrac{x^2}{2}- \dfrac{x^3}{3} - \dfrac{x^4}{4}-......)$
or, $= (-\dfrac{1}{x}) \times \ln (1-x) $
or, $=-\ln \dfrac{(1-x)}{x}; -1 \leq x \leq 1$
