If int frac{x^2+1}{x^4+1} dx = f(x) + c,then | vedclass

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(B) To solve the integral $I = \int \frac{x^2+1}{x^4+1} dx$,we divide the numerator and denominator by $x^2$: $I = \int \frac{1 + \frac{1}{x^2}}{x^2 +

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$\frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{x^2-1}{\sqrt{2}x}ight)$

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(B) To solve the integral $I = \int \frac{x^2+1}{x^4+1} dx$,we divide the numerator and denominator by $x^2$: $I = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$ We can rewrite the denominator as $(x - \frac{1}{x})^2 + 2$: $I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + (\sqrt{2})^2} dx$ Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) dx$. Substituting these into the integral: $I = \int \frac{dt}{t^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{t}{\sqrt{2}}ight) + c$ Substituting $t = x - \frac{1}{x} = \frac{x^2-1}{x}$ back: $I = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{x^2-1}{\sqrt{2}x}ight) + c$ Thus,$f(x) = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{x^2-1}{\sqrt{2}x}ight)$.

If $\int \frac{x^2+1}{x^4+1} dx = f(x) + c$,then $f(x)$ is equal to

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