int frac{x^2+1}{x^4-x^2+1} , dx = | vedclass

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(D) 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}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) 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 $(x - \frac{1}{x})^2 + 1$: $I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 1} \, dx$. Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) \, dx$. Substituting these into the integral: $I = \int \frac{dt}{t^2 + 1} = 	an^{-1}(t) + c$. Substituting $t = x - \frac{1}{x}$ back: $I = 	an^{-1}\left(x - \frac{1}{x}ight) + c = 	an^{-1}\left(\frac{x^2-1}{x}ight) + c$.

$\int \frac{x^2+1}{x^4-x^2+1} \, dx =$

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