int frac{x^{2}+1}{x^{4}+x^{2}+1} d x= | vedclass

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

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}} 	an ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}ight)+c$

Vedclass · Answer

(A) Let $I = \int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$. Divide the numerator and denominator by $x^{2}$: $I = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} dx$. Rewrite the denominator as $(x^{2} + \frac{1}{x^{2}} - 2) + 3 = (x - \frac{1}{x})^{2} + (\sqrt{3})^{2}$. So,$I = \int \frac{1 + \frac{1}{x^{2}}}{(x - \frac{1}{x})^{2} + (\sqrt{3})^{2}} dx$. Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^{2}}) dx$. Substituting these into the integral,we get: $I = \int \frac{dt}{t^{2} + (\sqrt{3})^{2}}$. Using the standard integral formula $\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$: $I = \frac{1}{\sqrt{3}} 	an^{-1}(\frac{t}{\sqrt{3}}) + c$. Replacing $t$ with $x - \frac{1}{x}$,we get: $I = \frac{1}{\sqrt{3}} 	an^{-1}(\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$.

$\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x=$

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