Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}& {\int }_{\alpha }^{\log 4}\frac{\mathrm{dx}}{\sqrt{{\mathrm{e}}^{\mathrm{x}}-1}}=\frac{\pi }{6}\\ & \text{ Let }{\mathrm{e}}^{\mathrm{x}}-1={\mathrm{t}}^2\\ & {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=2\mathrm{t}\mathrm{dt}\\ & =\int \frac{2\mathrm{dt}}{{\mathrm{t}}^2+1}\\ & =2{\tan }^{-1}\mathrm{t}\\ & ={2{\tan }^{-1}\left(\sqrt{{\mathrm{e}}^{\mathrm{x}}-1}\right)|}_{\alpha }^{{\log }_{\mathrm{e}}^4}\\ & =2\left[{\tan }^{-1}\sqrt{3}-{\tan }^{-1}\sqrt{{\mathrm{e}}^{\alpha }-1}\right]=\frac{\pi }{6}\\ & =\frac{\pi }{3}-{\tan }^{-1}\sqrt{{\mathrm{e}}^{\alpha }-1}=\frac{\pi }{12}\\ & \Rightarrow {\tan }^{-1}\sqrt{{\mathrm{e}}^{\alpha }-1}=\frac{\pi }{4}\\ & {\mathrm{e}}^{\alpha }=2{\mathrm{e}}^{-\alpha }=\frac{1}{2}\\ & {\mathrm{x}}^2-\left(2+\frac{1}{2}\right)\mathrm{x}+1=0\\ & 2{\mathrm{x}}^2-5\mathrm{x}+2=0\end{array}$