int_0^1 frac{x^4+1}{x^6+1} dx = | vedclass

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(A) To evaluate the integral $I = \int_0^1 \frac{x^4+1}{x^6+1} dx$,we can factor the denominator as $x^6+1 = (x^2+1)(x^4-x^2+1)$.However,a more effect

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$\frac{\pi}{3}$

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(A) To evaluate the integral $I = \int_0^1 \frac{x^4+1}{x^6+1} dx$,we can factor the denominator as $x^6+1 = (x^2+1)(x^4-x^2+1)$.However,a more effective method is to divide the numerator and denominator by $x^2$ or use partial fractions.Alternatively,we can write $x^4+1 = (x^4-x^2+1) + x^2$. Then $I = \int_0^1 \frac{x^4-x^2+1}{x^6+1} dx + \int_0^1 \frac{x^2}{x^6+1} dx$.$I = \int_0^1 \frac{1}{x^2+1} dx + \int_0^1 \frac{x^2}{(x^3)^2+1} dx$.For the first part,$\int_0^1 \frac{1}{x^2+1} dx = [	an^{-1}(x)]_0^1 = 	an^{-1}(1) - 	an^{-1}(0) = \frac{\pi}{4}$.For the second part,let $u = x^3$,then $du = 3x^2 dx$,so $x^2 dx = \frac{du}{3}$.When $x=0, u=0$ and when $x=1, u=1$.So,$\int_0^1 \frac{x^2}{(x^3)^2+1} dx = \frac{1}{3} \int_0^1 \frac{1}{u^2+1} du = \frac{1}{3} [	an^{-1}(u)]_0^1 = \frac{1}{3} (\frac{\pi}{4} - 0) = \frac{\pi}{12}$.Adding both parts,$I = \frac{\pi}{4} + \frac{\pi}{12} = \frac{3\pi + \pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.

$\int_0^1 \frac{x^4+1}{x^6+1} dx = $

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