intlimits_1^{sqrt 2 } {frac{{{x^2} + 1}}{{{x^ | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} 	an^{-1} \frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) Let $I = \int\limits_1^{\sqrt 2 } {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx$.Divide numerator and denominator by $x^2$:$I = \int\limits_1^{\sqrt 2 } {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \,dx}$.Rewrite the denominator as $(x - \frac{1}{x})^2 + 2$:$I = \int\limits_1^{\sqrt 2 } {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{(x - \frac{1}{x})}^2} + 2}}} \,dx$.Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^2}) dx$.When $x = 1$,$t = 1 - 1 = 0$.When $x = \sqrt{2}$,$t = \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{2-1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.Substituting these into the integral:$I = \int\limits_0^{1/\sqrt{2}} {\frac{1}{{{t^2} + 2}}} \,dt$.Using the formula $\int \frac{1}{t^2 + a^2} dt = \frac{1}{a} 	an^{-1}(\frac{t}{a}) + C$:$I = [\frac{1}{\sqrt{2}} 	an^{-1}(\frac{t}{\sqrt{2}})]_0^{1/\sqrt{2}}$.$I = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{1/\sqrt{2}}{\sqrt{2}}) - \frac{1}{\sqrt{2}} 	an^{-1}(0)$.$I = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{1}{2}) - 0 = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{1}{2})$.

$\int\limits_1^{\sqrt 2 } {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx$ is equal to:

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