Solution
= Solution
Multiply the <Taylor expansions> $\ln(1+x)=x-x^2/2+x^3/3+O(x^4)$ and $(1-x)^{-1}=1+x+x^2+O(x^3)$. Collecting the first three nonzero coefficients gives
$$
\boxed{\frac{\ln(1+x)}{1-x}=x+\frac{x^2}{2}+\frac{5x^3}{6}+O(x^4).}
$$
The cubic coefficient is $1-1/2+1/3=5/6$.