We have

$\log (1+x+x^2 + x^3) = \log \big [ (1+x)(1+x^2) \big ]$

$ \therefore \log (1+x+x^2+x^3) = \log \left [ \dfrac {(1+x)(1+x^2)(1-x)}{(1-x)} \right ]$

$ \therefore \log (1+x+x^2+x^3) = \log \left [ \dfrac {(1-x^4)}{(1-x)} \right ]$

$\therefore \log (1+x+x^2+x^3) = \log (1-x^4)- \log (1-x)$

$\therefore \log (1+x+x^2 +x^3) = \left ( -x^4 - \dfrac {x^8}{2} - \cdots \right ) - \left ( -x - \dfrac {x^2}{2} - \dfrac {x^3}{3} - \dfrac {x^4}{4} - \dfrac {x^5}{5}- \dfrac {x^6} {6} - \dfrac {x^7}{7} - \dfrac {x^8}{8}- \cdots \ \cdots \right ) $

$= x+ \dfrac {x^2}{2}+ \dfrac {x^3}{3} - \dfrac {3x^4}{4} + \dfrac {x^5}{5} + \dfrac {x^6}{6} + \dfrac {x^7}{7} - \dfrac {3x^8}{8} + \cdots \ \cdots $

$ \therefore \log (1+x+x^2 +x^3) = x + \dfrac {x^2} {2} + \dfrac {x^3}{3} - \dfrac {3x^4}{4} + \dfrac {x^5}{5} + \dfrac {x^6}{6} + \dfrac {x^7}{7} - \dfrac {3x^8}{8} + \cdots \ \cdots $