MathematicsMedium210×since 2002Q3406The value of the integral ∫−11loge(1−x+1+x)dx\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx}−1∫1loge(1−x+1+x)dx is equal to:A12loge2+π4−32{1 \over 2}{\log _e}2 + {\pi \over 4} - {3 \over 2}21loge2+4π−23B2loge2+π4−12{\log _e}2 + {\pi \over 4} - 12loge2+4π−1Cloge2+π2−1{\log _e}2 + {\pi \over 2} - 1loge2+2π−1D2loge2+π2−122{\log _e}2 + {\pi \over 2} - {1 \over 2}2loge2+2π−21Check answerSkip