Let y = $ log(1+x+x^2+x^3) \\ \; \\ = log\Bigg[ \dfrac{\big(1+x+x^2+x^3\big)\big(1-x\big)}{\big(1-x\big)} \Bigg] \; = \; log\bigg[\dfrac{ 1-x^4}{1-x} \bigg] \; = \; log(1-x^4) \; - \; log(1-x) \\ \; \\ \; \\ Using \; Standard \; expansions \; of \; log(1-x) \\ \; \\ \; \\ \therefore y= \Bigg[ -x^4 - \frac{x^8}{2}-\ldots \Bigg] \; - \; \Bigg[ -x - \frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\frac{x^6}{6}-\frac{x^7}{7}-\frac{x^8}{8}-\ldots \Bigg] \\ \; \\ \; \\ \therefore y= x + \frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}+\frac{x^8}{8} \; - \; x^4 - \frac{x^8}{2} \ldots \\ \; \\ \; \\ PROOF \; : \; Expansion \; of \; log(1-x) \\ \; \\ w=log(1-x) \\ \; \\ \therefore \frac{dw}{dx}=\frac{-1}{1-x}= \; -(1+x+x^2+\ldots) = \; -1-x-x^2-\ldots \\ \; \\ w = \; -x - \frac{x^2}{2}-\frac{x^3}{3}-\ldots \\ \; \\ \; \\ Now \; back, \\ y= x + \frac{x^2}{2}+\frac{x^3}{3}-\frac{3x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}+\frac{x^8}{8} - \frac{3x^8}{8} \ldots \\ \; \\ \; \\ log(1+x+x^2+x^3) = x + \frac{x^2}{2}+\frac{x^3}{3}-\frac{3x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}+\frac{x^8}{8} - \frac{3x^8}{8} \ldots $