int_{log _e 2}^x frac{d t}{sqrt{e^t-1}}=frac{ | vedclass

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Question

(A) Given integral: $\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6}$.Let $e^t - 1 = u^2$,then $e^t dt = 2u du$,so $dt = \frac{2u du}{u^2+1}

Vedclass · Accepted Answer

$2 \cdot \log _e 2$

Vedclass · Answer

(A) Given integral: $\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6}$.Let $e^t - 1 = u^2$,then $e^t dt = 2u du$,so $dt = \frac{2u du}{u^2+1}$.When $t = \log_e 2$,$u = \sqrt{e^{\log_e 2} - 1} = \sqrt{2-1} = 1$.When $t = x$,$u = \sqrt{e^x - 1}$.The integral becomes $\int_{1}^{\sqrt{e^x-1}} \frac{2u du}{(u^2+1)u} = 2 \int_{1}^{\sqrt{e^x-1}} \frac{du}{u^2+1} = 2 [	an^{-1} u]_{1}^{\sqrt{e^x-1}} = \frac{\pi}{6}$.$2 [	an^{-1}(\sqrt{e^x-1}) - 	an^{-1}(1)] = \frac{\pi}{6}$.$	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 \Rightarrow e^x = 4$.$x = \log_e 4 = \log_e(2^2) = 2 \log_e 2$.Thus,option $A$ is correct.

$\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6} \Rightarrow x=$

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