The value of the integral int limits_1^{sqrt{ | vedclass

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(B) Let $I = \int \limits_1^{\sqrt{2}+1} \frac{x^2-1}{x^2+1} \cdot \frac{1}{\sqrt{1+x^4}} \, dx$. Divide the numerator and denominator by $x^2$: $I =

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$\frac{\pi}{12 \sqrt{2}}$

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(B) Let $I = \int \limits_1^{\sqrt{2}+1} \frac{x^2-1}{x^2+1} \cdot \frac{1}{\sqrt{1+x^4}} \, dx$. Divide the numerator and denominator by $x^2$: $I = \int \limits_1^{\sqrt{2}+1} \frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \cdot \frac{1}{\sqrt{\frac{1}{x^2} + x^2}} \, dx = \int \limits_1^{\sqrt{2}+1} \frac{1 - \frac{1}{x^2}}{\sqrt{(x + \frac{1}{x})^2 - 2}} \cdot \frac{1}{x^2 + 1} \, dx$. Wait,let us rewrite the integral as: $I = \int \limits_1^{\sqrt{2}+1} \frac{1 - \frac{1}{x^2}}{\sqrt{x^2 + \frac{1}{x^2}}} \cdot \frac{1}{x^2 + 1} \, dx$ is not correct. Correct approach: $I = \int \limits_1^{\sqrt{2}+1} \frac{1 - \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \cdot \frac{x^2}{\sqrt{x^4+1}} \, dx$. Let $x + \frac{1}{x} = t$,then $(1 - \frac{1}{x^2}) \, dx = dt$. Also $x^2 + \frac{1}{x^2} = t^2 - 2$. For $x=1$,$t=2$. For $x=\sqrt{2}+1$,$t = \sqrt{2}+1 + \frac{1}{\sqrt{2}+1} = \sqrt{2}+1 + \sqrt{2}-1 = 2\sqrt{2}$. $I = \int \limits_2^{2\sqrt{2}} \frac{dt}{\sqrt{t^2-2} \cdot t} = \int \limits_2^{2\sqrt{2}} \frac{dt}{t \sqrt{t^2 - (\sqrt{2})^2}}$. Using the formula $\int \frac{dx}{x \sqrt{x^2-a^2}} = \frac{1}{a} \sec^{-1}(\frac{x}{a}) + C$: $I = \left[ \frac{1}{\sqrt{2}} \sec^{-1}(\frac{t}{\sqrt{2}}) \right]_2^{2\sqrt{2}} = \frac{1}{\sqrt{2}} [\sec^{-1}(2) - \sec^{-1}(\sqrt{2})] = \frac{1}{\sqrt{2}} [\frac{\pi}{3} - \frac{\pi}{4}] = \frac{1}{\sqrt{2}} [\frac{\pi}{12}] = \frac{\pi}{12\sqrt{2}}$.

The value of the integral $\int \limits_1^{\sqrt{2}+1} \left( \frac{x^2-1}{x^2+1} \right) \frac{1}{\sqrt{1+x^4}} \, dx$ is

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