Question

The value of the integral $\underset{1}{\overset{2}{\int }}\left(\frac{t^4+1}{t^6+1}\right)dt$ is :

• A. ${\tan }^{-1}\frac{1}{2}+\frac{1}{3}{\tan }^{-1}8-\frac{\pi }{3}$
• B. ${\tan }^{-1}2-\frac{1}{3}{\tan }^{-1}8+\frac{\pi }{3}$
• C. ${\tan }^{-1}2+\frac{1}{3}{\tan }^{-1}8-\frac{\pi }{3}$
• D. ${\tan }^{-1}\frac{1}{2}-\frac{1}{3}{\tan }^{-1}8+\frac{\pi }{3}$

Answer 1

Answer: (C)

$I=\underset{1}{\overset{2}{\int }}\left(\frac{t^4+1}{t^6+1}\right)dt$
$=\underset{1}{\overset{2}{\int }}\frac{\left(t^4+1-t^2\right)+t^2}{\left(t^2+1\right)\left(t^4-t^2+1\right)}dt$
$=\underset{1}{\overset{2}{\int }}\left(\frac{1}{t^2+1}+\frac{t^2}{t^6+1}\right)dt$
$=\underset{1}{\overset{2}{\int }}\left(\frac{1}{t^2+1}+\frac{1}{3}\frac{3t^2}{{\left(t^3\right)}^2+1}\right)dt$
$={\tan }^{-1}(t)+{\frac{1}{3}{\tan }^{-1}\left(t^3\right)|}_1^2$
$=\left({\tan }^{-1}(2)-{\tan }^{-1}(1)\right)+\frac{1}{3}\left({\tan }^{-1}\left(2^3\right)-{\tan }^{-1}\left(1^3\right)\right)$
$={\tan }^{-1}(2)+\frac{1}{3}{\tan }^{-1}(8)-\frac{\pi }{3}$