Let, $ y = log( 1-x+x^2 ) \; = \; log\bigg[ \dfrac{\big(1-x+x^2\big)\big(1+x\big)}{\big(1+x\big)} \bigg] \\ \; \\ y = log\bigg[ \dfrac{(1+x^3)}{1+x} \bigg] \; = \; log(1+x^3) \; - \; log(1+x) \\ \; \\ \; \\ Using \; Standard \; expansions \; of \; log(1+x) \; = \; x - \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \\ \; \\ \; \\ y=log(1+x^3) \; - \; log(1+x) \\ \; \\ = x^3 - \frac{x^6}{2}+\ldots-\bigg( x - \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \bigg) \\ \; \\ \; \\ = x^3 - \frac{x^6}{2}+\ldots- x + \frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}-\ldots \\ \; \\ \; \\ = -x + \frac{x^2}{2}+\frac{2x^3}{3}+\frac{x^4}{4}-\ldots \\ \; \\ \; \\ \therefore log( 1-x+x^2 ) \; = \; -x + \frac{x^2}{2} + \frac{2x^3}{3} +\frac{x^4}{4}-\ldots $