If int frac{1+x^2}{1+x^4} dx=frac{1}{sqrt{2}} | vedclass

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

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$x-\frac{1}{x}$

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(D) Let $I = \int \frac{1+x^2}{1+x^4} dx$.Divide the numerator and denominator by $x^2$:$I = \int \frac{\frac{1}{x^2} + 1}{\frac{1}{x^2} + x^2} dx = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$.We can write the denominator as $(x - \frac{1}{x})^2 + 2$:$I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 2} dx$.Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) dx$.Substituting these into the integral:$I = \int \frac{dt}{t^2 + 2} = \int \frac{dt}{t^2 + (\sqrt{2})^2}$.Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$:$I = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{t}{\sqrt{2}}) + c$.Substituting $t = x - \frac{1}{x}$ back:$I = \frac{1}{\sqrt{2}} 	an^{-1}\left[\frac{x - \frac{1}{x}}{\sqrt{2}}ight] + c$.Comparing this with the given expression $\frac{1}{\sqrt{2}} 	an^{-1}\left[\frac{f(x)}{\sqrt{2}}ight] + c$,we get $f(x) = x - \frac{1}{x}$.

If $\int \frac{1+x^2}{1+x^4} dx=\frac{1}{\sqrt{2}} \tan ^{-1}\left[\frac{f(x)}{\sqrt{2}}\right]+c$,then $f(x)=$

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