MathematicsMedium210×since 2002Q3453The value of the integral ∫12(t4+1t6+1)dt\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt}∫12(t6+1t4+1)dt isAtan−112−13tan−18+π3{\tan ^{ - 1}}{1 \over 2} - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}tan−121−31tan−18+3πBtan−12−13tan−18+π3{\tan ^{ - 1}}2 - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}tan−12−31tan−18+3πCtan−12+13tan−18−π3{\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}tan−12+31tan−18−3πDtan−112+13tan−18−π3{\tan ^{ - 1}}{1 \over 2} + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}tan−121+31tan−18−3πCheck answerSkip