int_1^2 frac{x^4-1}{x^6-1} d x= | vedclass

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(A) Let $I = \int_1^2 \frac{x^4-1}{x^6-1} d x$. We simplify the integrand: $\frac{x^4-1}{x^6-1} = \frac{(x^2-1)(x^2+1)}{(x^2-1)(x^4+x^2+1)} = \frac{x^

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$\frac{1}{\sqrt{3}} 	an ^{-1}\left(\frac{\sqrt{3}}{2}ight)$

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(A) Let $I = \int_1^2 \frac{x^4-1}{x^6-1} d x$. We simplify the integrand: $\frac{x^4-1}{x^6-1} = \frac{(x^2-1)(x^2+1)}{(x^2-1)(x^4+x^2+1)} = \frac{x^2+1}{x^4+x^2+1}$. Note that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$. Using partial fractions: $\frac{x^2+1}{(x^2+x+1)(x^2-x+1)} = \frac{1}{2} \left( \frac{1}{x^2+x+1} + \frac{1}{x^2-x+1} ight)$. Integrating: $I = \frac{1}{2} \int_1^2 \left( \frac{1}{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} + \frac{1}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} ight) d x$. $I = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \left[ 	an^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}ight) + 	an^{-1}\left(\frac{x-1/2}{\sqrt{3}/2}ight) ight]_1^2$. $I = \frac{1}{\sqrt{3}} \left[ 	an^{-1}\left(\frac{2x+1}{\sqrt{3}}ight) + 	an^{-1}\left(\frac{2x-1}{\sqrt{3}}ight) ight]_1^2$. Evaluating at limits: $I = \frac{1}{\sqrt{3}} \left[ (	an^{-1}(\frac{5}{\sqrt{3}}) + 	an^{-1}(\sqrt{3})) - (	an^{-1}(\sqrt{3}) + 	an^{-1}(\frac{1}{\sqrt{3}})) ight]$. $I = \frac{1}{\sqrt{3}} \left[ 	an^{-1}(\frac{5}{\sqrt{3}}) - 	an^{-1}(\frac{1}{\sqrt{3}}) ight]$. Using $	an^{-1} A - 	an^{-1} B = 	an^{-1}(\frac{A-B}{1+AB})$: $I = \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{5/\sqrt{3} - 1/\sqrt{3}}{1 + 5/3} ight) = \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{4/\sqrt{3}}{8/3} ight) = \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{\sqrt{3}}{2} ight)$.

$\int_1^2 \frac{x^4-1}{x^6-1} d x=$

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