Solution
= Solution
Expand the <natural logarithm> and reciprocal separately:
$$
\ln(2+x)=\ln2+\frac x2-\frac{x^2}{8}+O(x^3),\qquad
\frac1{2-x}=\frac12\left(1+\frac x2+\frac{x^2}{4}+O(x^3)\right).
$$
Multiplication gives the first three nonzero terms of the <Maclaurin series>:
$$
\boxed{\frac{\ln(2+x)}{2-x}=\frac{\ln2}{2}
+\frac{1+\ln2}{4}x+\frac{1+2\ln2}{16}x^2+O(x^3).}
$$
The local expansions converge for $|x|<2$.