• # question_answer In  the expansion of ${{\log }_{e}}\frac{1}{1-x-{{x}^{2}}+{{x}^{3}}}$,  the coefficient of $x$ is A) 0 B) 1 C) ? 1 D) 1/2

${{\log }_{e}}\left[ \frac{1}{1-x-{{x}^{2}}+{{x}^{3}}} \right]={{\log }_{e}}\left[ \frac{1}{(1-x)-{{x}^{2}}(1-x)} \right]$ $={{\log }_{e}}\left[ \frac{1}{(1-x)(1-{{x}^{2}})} \right]={{\log }_{e}}\left[ \frac{1}{{{(1-x)}^{2}}(1+x)} \right]$ $={{\log }_{e}}\{{{(1-x)}^{-2}}{{(1+x)}^{-1}}\}$ ??(i) $={{\log }_{e}}{{(1-x)}^{-2}}+{{\log }_{e}}{{(1+x)}^{-1}}$ $=-2{{\log }_{e}}(1-x)-{{\log }_{e}}(1+x)$ $=-2\,\left[ -x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}-.......\infty \right]$                 $-\left[ x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+.....\infty \right]$, $(\because \ -1<x\le 1)$ Hence coefficient of $x=2-1=1$.