The value of the integral \int_{1}^{2} (\frac{t^4+1}{t^6+1}) \dt is : [29-Jan-2023 Shift 2]

Correct Answer is : tan−12+13tan−18−π3

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

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

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

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

JEE Mains 29-Jan-2023 Shift 2 Solved Paper

Section: Mathematics

Question : 65 of 90

Marks: +1, -0

The value of the integral

∫

(‌

t4+1

t6+1

)‌dt is :

[29-Jan-2023 Shift 2]

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