$\log (1+y)= y-\dfrac {y^2}{2}+ \dfrac {y^3}{3} \cdots \ \cdots \ Put \ y=-x+x^2 \ \log (1-x+x^2)$ ***$*$= -x+x^2 - \dfrac {(-x+x^2)^2}{2}+ \dfrac {(-x+x^2)^3}{3}- \cdots \ \cdots (higher \ exponents \ of \ x) $

$= -x+x^2 - \dfrac {x^4-2x^3+x^2}{2}+ \dfrac {x^6-3x^5 +3x^4-x^3}{3} - \cdots$

$ = -x+x^2-\dfrac {x^2}{2}+ \dfrac {2x^3}{2}- \dfrac {x^3}{3}+ \cdots \ \cdots (higher \ exponents \ of \ x) $

$= -x+\dfrac {x^2}{2}+ x^3 \left ( \dfrac 1- \dfrac {1}{3} \right )+ \cdots \ \cdots $

$\log (1-x+x^2)$

$\log (1-x+x^2)= - x + \dfrac {x^2}{2}+ \dfrac {2x^3}{3}+ \cdots \ \cdots$