If $\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c} $, $g(1) = 0$, then $g\left( {{1 \over 2}} \right)$ is equal to :

Question

${\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) + {\pi \over 3}$

${\log _e}\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right) + {\pi \over 3}$

${\log _e}\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right) - {\pi \over 3}$

${1 \over 2}{\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) - {\pi \over 6}$

Please login to view the detailed solution steps...