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

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Aicharya · Accepted Answer

$\begin{aligned} & \int_\alpha^{\log 4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{x}}-1}}=\frac{\pi}{6} \ & 	ext { Let } \mathrm{e}^{\mathrm{x}}-1=\mathrm{t}^2 \ & \mathrm{e}^{\mathrm{x}} \mathrm{dx}=2 \mathrm{t} \mathrm{dt} \ & =\int \frac{2 \mathrm{dt}}{\mathrm{t}^2+1} \ & =2 	an ^{-1} \mathrm{t} \ & =\left.2 	an ^{-1}\left(\sqrt{\mathrm{e}^{\mathrm{x}}-1}ight)ight|_\alpha ^{\log _{\mathrm{e}}^4} \ & =2\left[	an ^{-1} \sqrt{3}-	an ^{-1} \sqrt{\mathrm{e}^\alpha-1}ight]=\frac{\pi}{6} \ & =\frac{\pi}{3}-	an ^{-1} \sqrt{\mathrm{e}^\alpha-1}=\frac{\pi}{12} \ & \Rightarrow 	an ^{-1} \sqrt{\mathrm{e}^\alpha-1}=\frac{\pi}{4} \ & \mathrm{e}^\alpha=2 \quad \mathrm{e}^{-\alpha}=\frac{1}{2} \ & \mathrm{x}^2-\left(2+\frac{1}{2}ight) \mathrm{x}+1=0 \ & 2 \mathrm{x}^2-5 \mathrm{x}+2=0\end{aligned}$

Let $\int_\alpha^{\log _e 4} \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6}$. Then…

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