If A=int_0^{infty} frac{1+x^2}{1+x^4} d x and | vedclass

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(C) Let $I = \int \frac{1+x^2}{1+x^4} dx = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$. For $A = \int_0^{\infty} \frac{1+x^2}{1+x^4} dx$,we

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$2B=A$

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(C) Let $I = \int \frac{1+x^2}{1+x^4} dx = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$. For $A = \int_0^{\infty} \frac{1+x^2}{1+x^4} dx$,we divide numerator and denominator by $x^2$: $A = \int_0^{\infty} \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 2} dx$. Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^2}) dx$. As $x 	o 0, t 	o -\infty$ and as $x 	o \infty, t 	o \infty$. $A = \int_{-\infty}^{\infty} \frac{dt}{t^2 + 2} = \left[ \frac{1}{\sqrt{2}} 	an^{-1}(\frac{t}{\sqrt{2}}) ight]_{-\infty}^{\infty} = \frac{1}{\sqrt{2}} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{\pi}{\sqrt{2}}$. For $B = \int_0^1 \frac{1+x^2}{1+x^4} dx$,let $x = \frac{1}{u}$,then $dx = -\frac{1}{u^2} du$. $B = \int_{\infty}^1 \frac{1 + \frac{1}{u^2}}{1 + \frac{1}{u^4}} (-\frac{1}{u^2}) du = \int_1^{\infty} \frac{u^2 + 1}{u^4 + 1} du$. Since $A = \int_0^1 \frac{1+x^2}{1+x^4} dx + \int_1^{\infty} \frac{1+x^2}{1+x^4} dx = B + B = 2B$. Therefore,$2B = A$.

If $A=\int_0^{\infty} \frac{1+x^2}{1+x^4} d x$ and $B=\int_0^1 \frac{1+x^2}{1+x^4} d x$,then

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