The value of int frac{x^{2}+1}{x^{4}+x^{2}+1} | vedclass

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Question

(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 + 1/x^{2}}{x^{2} + 1 + 1/x^{2}

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}} 	an ^{-1}\left\{\frac{x-1/x}{\sqrt{3}}ight\}+C$

Vedclass · Answer

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

The value of $\int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$ is

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