The value of int_0^1 frac{x^4 + 1}{x^2 + 1} , | vedclass

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(A) Let $I = \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$. We can rewrite the integrand as $\frac{x^4 - 1 + 2}{x^2 + 1} = \frac{(x^2 - 1)(x^2 + 1) + 2}{x^2

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$\frac{1}{6}(3\pi - 4)$

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(A) Let $I = \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$. We can rewrite the integrand as $\frac{x^4 - 1 + 2}{x^2 + 1} = \frac{(x^2 - 1)(x^2 + 1) + 2}{x^2 + 1} = x^2 - 1 + \frac{2}{x^2 + 1}$. Now,integrate term by term: $I = \int_0^1 (x^2 - 1) \, dx + 2 \int_0^1 \frac{1}{x^2 + 1} \, dx$. Evaluating the integrals: $I = \left[ \frac{x^3}{3} - x ight]_0^1 + 2 \left[ 	an^{-1} x ight]_0^1$. $I = \left( \frac{1}{3} - 1 ight) - (0) + 2 \left( 	an^{-1}(1) - 	an^{-1}(0) ight)$. $I = -\frac{2}{3} + 2 \left( \frac{\pi}{4} - 0 ight)$. $I = -\frac{2}{3} + \frac{\pi}{2} = \frac{3\pi - 4}{6}$.

The value of $\int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$ is

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