The value of the integral $\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt} $ is

Question

${\tan ^{ - 1}}{1 \over 2} - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}$

${\tan ^{ - 1}}2 - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}$

${\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$

${\tan ^{ - 1}}{1 \over 2} + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$

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