If int_{log 2}^x frac{du}{({e^u} - 1)^{1/2}} | vedclass

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(C) Let $I = \int_{\log 2}^x \frac{du}{({e^u} - 1)^{1/2}} = \frac{\pi}{6}$.Substitute $e^u - 1 = t^2$,so $e^u du = 2t dt$,which implies $du = \frac{2t

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$4$

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(C) Let $I = \int_{\log 2}^x \frac{du}{({e^u} - 1)^{1/2}} = \frac{\pi}{6}$.Substitute $e^u - 1 = t^2$,so $e^u du = 2t dt$,which implies $du = \frac{2t}{e^u} dt = \frac{2t}{t^2 + 1} dt$.When $u = \log 2$,$t = \sqrt{e^{\log 2} - 1} = \sqrt{2 - 1} = 1$.When $u = x$,$t = \sqrt{e^x - 1}$.The integral becomes $\int_{1}^{\sqrt{e^x - 1}} \frac{2t}{t^2 + 1} \cdot \frac{1}{t} dt = \int_{1}^{\sqrt{e^x - 1}} \frac{2}{t^2 + 1} dt = \frac{\pi}{6}$.$2 [	an^{-1} t]_{1}^{\sqrt{e^x - 1}} = \frac{\pi}{6}$.$	an^{-1}(\sqrt{e^x - 1}) - 	an^{-1}(1) = \frac{\pi}{12}$.$	an^{-1}(\sqrt{e^x - 1}) - \frac{\pi}{4} = \frac{\pi}{12}$.$	an^{-1}(\sqrt{e^x - 1}) = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi + 3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.$\sqrt{e^x - 1} = 	an(\frac{\pi}{3}) = \sqrt{3}$.$e^x - 1 = 3$.$e^x = 4$.

If $\int_{\log 2}^x \frac{du}{({e^u} - 1)^{1/2}} = \frac{\pi}{6}$,then ${e^x} = $

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