The value of int frac{x^{2}-1}{x^{4}+3 x^{2}+ | vedclass

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(A) Let $I = \int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$. Divide the numerator and denominator by $x^{2}$: $I = \int \frac{1 - 1/x^{2}}{x^{2} + 3 + 1/x^

Vedclass · Accepted Answer

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

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

The value of $\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$ for $x>0$ is

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