int frac{x^2+1}{x^4+7 x^2+1} d x is equal to | vedclass

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Question

(A) Let $I = \int \frac{x^2+1}{x^4+7 x^2+1} d x$. Divide the numerator and denominator by $x^2$: $I = \int \frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}}

Vedclass · Accepted Answer

$\frac{1}{3} 	an ^{-1}\left(\frac{x^2-1}{3 x}ight)+C$

Vedclass · Answer

(A) Let $I = \int \frac{x^2+1}{x^4+7 x^2+1} d x$. Divide the numerator and denominator by $x^2$: $I = \int \frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}} d x = \int \frac{1+\frac{1}{x^2}}{(x^2+\frac{1}{x^2}+2)+5} d x$. $I = \int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+9} d x$. Let $t = x - \frac{1}{x}$, then $dt = (1 + \frac{1}{x^2}) dx$. Substituting these into the integral: $I = \int \frac{dt}{t^2 + 3^2} = \frac{1}{3} 	an^{-1}(\frac{t}{3}) + C$. Substituting $t = x - \frac{1}{x}$ back: $I = \frac{1}{3} 	an^{-1}(\frac{x - \frac{1}{x}}{3}) + C = \frac{1}{3} 	an^{-1}(\frac{x^2-1}{3x}) + C$.

$\int \frac{x^2+1}{x^4+7 x^2+1} d x$ is equal to

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