Let, $ y=log(1+x) \\ \; \\ \dfrac{dy}{dx}=\dfrac{1}{1+x}= (1+x)^{-1} \\ \; \\ By \; using \; standard \; expansion \; of \; \dfrac{1}{1+x} \; , \\ \; \\ \therefore \dfrac{dy}{dx} \; = \; \; 1-x+x^2-x^3+\ldots \\ \{ \because Binomial \; Expansion \; : \; \dfrac{1}{1+x} = 1-x+x^2-x^3+\ldots \} \\ \; \\ On \; integrating, \; \\ \; \\ \therefore y \; = \; x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\ldots \\ \; \\ Now \; replace \; x \; by \; (x-1) \\ \; \\ \; \\ \therefore log\Big[ 1+x-1 \Big] \; = \; (x-1) - \dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}-\dfrac{(x-1)^4}{4}+\ldots \\ \; \\ \; \\ \therefore log\; x \; = \; (x-1) - \dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}-\dfrac{(x-1)^4}{4}+\ldots \ldots Hence ]; Proved $