• # question_answer If  $y=2{{x}^{2}}-1$, then  $\left[ \frac{1}{y}+\frac{1}{3{{y}^{3}}}+\frac{1}{5{{y}^{5}}}+.... \right]$ is equal to A) $\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}-..... \right]$ B) $\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}+..... \right]$ C) $\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}+..... \right]$ D) $\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}-..... \right]$

Given that y = 2x2?1, then $\left[ \frac{1}{y}+\frac{1}{3{{y}^{3}}}+\frac{1}{5{{y}^{5}}}+.... \right]\,=\frac{1}{2}\left[ {{\log }_{e}}\left( \frac{1+\frac{1}{y}}{1-\frac{1}{y}} \right) \right]$ $=\frac{1}{2}{{\log }_{e}}\left( \frac{y+1}{y-1} \right)=-\frac{1}{2}{{\log }_{e}}\left( \frac{y-1}{y+1} \right)$ $=-\frac{1}{2}{{\log }_{e}}\left( 1-\frac{2}{1+y} \right)=-\frac{1}{2}{{\log }_{e}}\left( 1-\frac{1}{{{x}^{2}}} \right)$ $=\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}+.... \right]$.