The value of the integral 1 2 ( t 4 + 1 t 6 + 1 ) d t is [2023]

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Q.
$The value of the integral \int_{1}^{2} (\frac{t^{4} + 1}{t^{6} + 1}) d t is$ [2023]

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

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

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

4 $\tan^{- 1} \frac{1}{2} - \frac{1}{3} \tan^{- 1} 8 + \frac{π}{3}$

Ans.
(2)

Let $I = \int_{1}^{2} [\frac{t^{4} + 1}{t^{6} + 1}] d t$

$= \int_{1}^{2} \frac{1}{t^{2} + 1} d t + \frac{1}{3} \int_{1}^{2} \frac{3 t^{2}}{{(t^{3})}^{2} + 1} d t$

$= {[\tan^{- 1} t]}_{1}^{2} + {[\frac{1}{3} (\tan^{- 1} t^{3})]}_{1}^{2}$

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

Solution

Q.
$The value of the integral \int_{1}^{2} (\frac{t^{4} + 1}{t^{6} + 1}) d t is$ [2023]

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

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

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

4 $\tan^{- 1} \frac{1}{2} - \frac{1}{3} \tan^{- 1} 8 + \frac{π}{3}$

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Solution

Q. The value of the integral ∫12(t4+1t6+1)dt is [2023]

1 tan-12-13tan-18+π3

2 tan-12+13tan-18-π3

3 tan-112+13tan-18-π3

4 tan-112-13tan-18+π3

Ans. (2) Let I=∫12[t4+1t6+1]dt =∫121t2+1dt+13∫123t2(t3)2+1dt =[tan-1t]12+[13(tan-1t3)]12 =tan-12+13tan-18-π3

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Q.
$The value of the integral \int_{1}^{2} (\frac{t^{4} + 1}{t^{6} + 1}) d t is$ [2023]

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

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

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

4 $\tan^{- 1} \frac{1}{2} - \frac{1}{3} \tan^{- 1} 8 + \frac{π}{3}$