The value of the integral $\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right) d t$ is :

(C) $\tan ^{-1} 2+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}$

(A) $\tan ^{-1} \frac{1}{2}+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}$

(B) $\tan ^{-1} 2-\frac{1}{3} \tan ^{-1} 8+\frac{\pi}{3}$

(D) $\tan ^{-1} \frac{1}{2}-\frac{1}{3} \tan ^{-1} 8+\frac{\pi}{3}$

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Definite Integrals

The value of the integral

\int_{1}^{2} (\frac{t ^{4} + 1}{t ^{6} + 1}) d t

is :

Answer: (C)

Split the integrand using

t^{6} + 1 = (t^{2} + 1) (t^{4} - t^{2} + 1)

and

t^{4} + 1 = (t^{4} - t^{2} + 1) + t^{2}

, which gives

\frac{t ^{4} + 1}{t ^{6} + 1} = \frac{1}{t ^{2} + 1} + \frac{t ^{2}}{t ^{6} + 1}

. The first part integrates to

tan^{- 1} t

, and the second, written as

\frac{1}{3} \cdot \frac{3 t ^{2}}{( t ^{3} ) ^{2} + 1}

, integrates to

\frac{1}{3} tan^{- 1} (t^{3})

. Between

1

and

2

\int_{1}^{2} \frac{d t}{t ^{2} + 1} = tan^{- 1} 2 - \frac{π}{4}

and

\frac{1}{3} [tan^{- 1} (t^{3})]_{1}^{2} = \frac{1}{3} tan^{- 1} 8 - \frac{π}{12}

. Adding, the value is

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

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option