Solution
= Solution
Use the real <logarithm> near zero and subtract its two <Taylor series>:
$$
\ln\frac{1-x}{1+2x^2}=\ln(1-x)-\ln(1+2x^2)=\left(-x-\frac{x^2}2-\frac{x^3}3\right)-2x^2+O(x^4).
$$
Thus the first three nonzero terms are
$$
\boxed{-x-\frac52x^2-\frac13x^3.}
$$