If $2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,$ then $\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$ is equalto :

Question

If $2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,$

then $\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$ is equalto :

log4

${\pi \over 2}$ + log2

log2

${\pi \over 2}$ $-$ log4

Solution

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