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Question

Mathematics Question on Definite Integral

The value of the integral ∫12(t4+1t6+1)\int^2_1\bigg(\frac{t^4+1}{t^6+1}\bigg)dt is

A

tan−12−13tan−18+π3tan^{-1}2-\frac{1}{3}tan^{-1}8+\frac{\pi}{3}

B

tan−12−13tan−18−π3tan^{-1}2-\frac{1}{3}tan^{-1}8-\frac{\pi}{3}

C

tan−112+13tan−18−π3tan^{-1}\frac{1}{2}+\frac{1}{3}tan^{-1}8-\frac{\pi}{3}

D

tan−112−13tan−18+π3tan^{-1}\frac{1}{2}-\frac{1}{3}tan^{-1}8+\frac{\pi}{3}

Answer

tan−12−13tan−18−π3tan^{-1}2-\frac{1}{3}tan^{-1}8-\frac{\pi}{3}

Explanation

Solution

The Correct Option is (B): tan−12−13tan−18−π3tan^{-1}2-\frac{1}{3}tan^{-1}8-\frac{\pi}{3}