The value of the integral _1^2(t^4+1)/(t^6+1),dt is:

Question

The value of the integral $\displaystyle\int_1^2\frac{t^4+1}{t^6+1}\,dt$ is:

Accepted Answer

$$\int_1^2 {{{{t^4} + 1} \over {{t^6} + 1}}dt} $$

$$ = \int_1^2 {{{{{({t^2} + 1)}^2}} \over {{t^6} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{t^6} + 1}}dt} } $$

$$ = \int_1^2 {{{{t^2} + 1} \over {{t^4} - {t^2} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{{({t^3})}^2} + 1}}dt} } $$

$$ = \left. {{{\tan }^{ - 1}}(2t + \sqrt 3 ) + {{\tan }^{ - 1}}(2t - \sqrt 3 )} \right|_1^2 - \left. {{2 \over 3}{{\tan }^{ - 1}}({t^3})} \right|_1^2$$

$$ = {\tan ^{ - 1}}(4 + \sqrt 3 ) + {\tan ^{ - 1}}(4 - \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2\sqrt 3 ) - {2 \over 3}({\tan ^{ - 1}}8 - {\tan ^{ - 1}}1)$$

$$ = {\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi  \over 3}$$

Solutions · 1