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