• question_answer The value of  ${{\log }_{e}}\left( 1+a{{x}^{2}}+{{a}^{2}}+\frac{a}{{{x}^{2}}} \right)$ is A) $a\,\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)-\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}-\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}-\frac{1}{{{x}^{6}}} \right)-.....$ B)  $a\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)-\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}+\frac{1}{{{x}^{6}}} \right)-.....$ C) $a\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)+\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}+\frac{1}{{{x}^{6}}} \right)+.....$ D)  $a\,\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)+\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}-\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}-\frac{1}{{{x}^{6}}} \right)+.....$

${{\log }_{e}}\left[ 1+a{{x}^{2}}+{{a}^{2}}+\frac{a}{{{x}^{2}}} \right]$ $={{\log }_{e}}(1+a{{x}^{2}})\,\left( 1+\frac{a}{{{x}^{2}}} \right)$ $={{\log }_{e}}(1+a{{x}^{2}})+{{\log }_{e}}\left( 1+\frac{a}{{{x}^{2}}} \right)$ = $\left[ a{{x}^{2}}-\frac{1}{2}{{a}^{2}}{{x}^{4}}+\frac{1}{3}{{a}^{3}}{{x}^{6}}-........... \right]$ $+\left[ \frac{a}{{{x}^{2}}}-\frac{1}{2}{{a}^{2}}\left( \frac{1}{{{x}^{4}}} \right)+\frac{1}{3}{{a}^{3}}\left( \frac{1}{{{x}^{6}}} \right)-..... \right]$     $=a\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)-\frac{1}{2}.{{a}^{2}}\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+\frac{1}{3}{{a}^{3}}\left( {{x}^{6}}+\frac{1}{{{x}^{6}}} \right)-...$