MathematicsMedium210×since 2002Q3353If 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 :Alog4Bπ2{\pi \over 2}2π + log2Clog2Dπ2{\pi \over 2}2π −-− log4Check answerSkip