Let int_alpha^{log _e 4} frac{dx}{sqrt{e^{x}- | vedclass

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(A) Given the integral $\int_\alpha^{\log _e 4} \frac{dx}{\sqrt{e^{x}-1}}=\frac{\pi}{6}$.Let $e^{x}-1=t^2$,then $e^x dx = 2t dt$,which implies $dx = \

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$2 x^2-5 x+2=0$

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(A) Given the integral $\int_\alpha^{\log _e 4} \frac{dx}{\sqrt{e^{x}-1}}=\frac{\pi}{6}$.Let $e^{x}-1=t^2$,then $e^x dx = 2t dt$,which implies $dx = \frac{2t dt}{t^2+1}$.When $x = \log_e 4$,$t = \sqrt{e^{\log_e 4}-1} = \sqrt{4-1} = \sqrt{3}$.When $x = \alpha$,$t = \sqrt{e^\alpha-1}$.The integral becomes $\int_{\sqrt{e^\alpha-1}}^{\sqrt{3}} \frac{2t dt}{t(t^2+1)} = 2 \int_{\sqrt{e^\alpha-1}}^{\sqrt{3}} \frac{dt}{t^2+1} = 2 [	an^{-1} t]_{\sqrt{e^\alpha-1}}^{\sqrt{3}}$.This equals $2(	an^{-1} \sqrt{3} - 	an^{-1} \sqrt{e^\alpha-1}) = 2(\frac{\pi}{3} - 	an^{-1} \sqrt{e^\alpha-1}) = \frac{\pi}{6}$.Dividing by $2$,we get $\frac{\pi}{3} - 	an^{-1} \sqrt{e^\alpha-1} = \frac{\pi}{12}$,so $	an^{-1} \sqrt{e^\alpha-1} = \frac{\pi}{3} - \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}$.Thus,$\sqrt{e^\alpha-1} = 	an(\frac{\pi}{4}) = 1$,which means $e^\alpha-1 = 1$,so $e^\alpha = 2$.Then $e^{-\alpha} = \frac{1}{2}$.The quadratic equation with roots $2$ and $\frac{1}{2}$ is $(x-2)(x-\frac{1}{2}) = 0$,which simplifies to $x^2 - \frac{5}{2}x + 1 = 0$,or $2x^2 - 5x + 2 = 0$.

Let $\int_\alpha^{\log _e 4} \frac{dx}{\sqrt{e^{x}-1}}=\frac{\pi}{6}$. Then $e^\alpha$ and $e^{-\alpha}$ are the roots of the equation :

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