JEE Main 20169 Apr, Morning shiftMathematicsMediumIntegral Calculus369× since 2002If 2∫01tan−1xdx=∫01cot−1(1−x+x2)dx,2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,20∫1tan−1xdx=0∫1cot−1(1−x+x2)dx, then ∫01tan−1(1−x+x2)dx\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx0∫1tan−1(1−x+x2)dx is equalto :AOption A. log4BOption B. π2{\pi \over 2}2π + log2COption C. log2DOption D. π2{\pi \over 2}2π −-− log4CheckSkip · show solution
