The value of the integral int limits_1^2 left | vedclass

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(C) Let $I = \int \limits_1^2 \left(\frac{t^4+1}{t^6+1}\right) dt$.We can rewrite the integrand as:$\frac{t^4+1}{t^6+1} = \frac{(t^4-t^2+1) + t^2}{(t^

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$	an ^{-1} 2+\frac{1}{3} 	an ^{-1} 8-\frac{\pi}{3}$

Vedclass · Answer

(C) Let $I = \int \limits_1^2 \left(\frac{t^4+1}{t^6+1}ight) dt$.We can rewrite the integrand as:$\frac{t^4+1}{t^6+1} = \frac{(t^4-t^2+1) + t^2}{(t^2+1)(t^4-t^2+1)} = \frac{1}{t^2+1} + \frac{t^2}{t^6+1}$.Now,integrate term by term:$I = \int \limits_1^2 \frac{1}{t^2+1} dt + \int \limits_1^2 \frac{t^2}{(t^3)^2+1} dt$.For the second integral,let $u = t^3$,then $du = 3t^2 dt$,so $t^2 dt = \frac{1}{3} du$.$I = [	an^{-1}(t)]_1^2 + \frac{1}{3} [	an^{-1}(t^3)]_1^2$.Evaluating the limits:$I = (	an^{-1}(2) - 	an^{-1}(1)) + \frac{1}{3} (	an^{-1}(8) - 	an^{-1}(1))$.Since $	an^{-1}(1) = \frac{\pi}{4}$:$I = 	an^{-1}(2) - \frac{\pi}{4} + \frac{1}{3} 	an^{-1}(8) - \frac{1}{3} \cdot \frac{\pi}{4}$.$I = 	an^{-1}(2) + \frac{1}{3} 	an^{-1}(8) - \frac{\pi}{4} - \frac{\pi}{12} = 	an^{-1}(2) + \frac{1}{3} 	an^{-1}(8) - \frac{4\pi}{12}$.$I = 	an^{-1}(2) + \frac{1}{3} 	an^{-1}(8) - \frac{\pi}{3}$.

The value of the integral $\int \limits_1^2 \left(\frac{t^4+1}{t^6+1}\right) dt$ is $..........$.

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