Definite Integration Question 12
Question 12 - 29 January - Shift 2
The value of the integral $\int_1^{2}(\frac{t^{4}+1}{t^{6}+1}) d t$ is
(1) $\tan ^{-1} \frac{1}{2}+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}$
(2) $\tan ^{-1} 2-\frac{1}{3} \tan ^{-1} 8+\frac{\pi}{3}$
(3) $\tan ^{-1} 2+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}$
(4) $\tan ^{-1} \frac{1}{2}-\frac{1}{3} \tan ^{-1} 8+\frac{\pi}{3}$
Show Answer
Answer: (3)
Solution:
Formula: Standard formulas for Indefinite Integration
$I=\int_1^{2}(\frac{t^{4}+1}{t^{6}+1}) d t$
$I=\int_1^{2} \frac{(t^{4}+1-t^{2})+t^{2}}{(t^{2}+1)(t^{4}-t^{2}+1)} d t$
$I=\int_1^{2}(\frac{1}{t^{2}+1}+\frac{t^{2}}{t^{6}+1}) d t$
$I=\int_1^{2}(\frac{1}{t^{2}+1}+\frac{1}{3} \frac{3 t^{2}}{(t^{3})^{2}+1}) d t$
$I=\tan ^{-1}(t)+.\frac{1}{3} \tan ^{-1}(t^{3})|_1 ^{2}$
$I=(\tan ^{-1}(2)-\tan ^{-1}(1))+\frac{1}{3}(\tan ^{-1}(2^{3})-\tan ^{-1}(1^{3}))$
$I=\tan ^{-1}(2)+\frac{1}{3} \tan ^{-1}(8)-\frac{\pi}{3}$
