JEE Main 202226 Jun, Evening shiftMathematicsMediumIntegral Calculus369× since 2002If ∫1x1−x1+xdx=g(x)+c\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c}∫x11+x1−xdx=g(x)+c, g(1)=0g(1) = 0g(1)=0, then g(12)g\left( {{1 \over 2}} \right)g(21) is equal to :AOption A. loge(3−13+1)+π3{\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) + {\pi \over 3}loge(3+13−1)+3πBOption B. loge(3+13−1)+π3{\log _e}\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right) + {\pi \over 3}loge(3−13+1)+3πCOption C. loge(3+13−1)−π3{\log _e}\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right) - {\pi \over 3}loge(3−13+1)−3πDOption D. 12loge(3−13+1)−π6{1 \over 2}{\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) - {\pi \over 6}21loge(3+13−1)−6πCheckSkip · show solution
