Let ∫_α^log _e 4 d x √ e^x-1= π 6. Then e^α and e^-α are the roots of the equation :

$$\begin{aligned} & \int_\alpha^{\log 4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{x}}-1}}=\frac{\pi}{6} \\ & \text { Let } \mathrm{e}^{\mathrm{x}}-1=\mathrm{t}^2 \\ & \mathrm{e}^{\mathrm{x}} \mathrm{dx}=2 \mathrm{t} \mathrm{dt} \\ & =\int \frac{2 \mathrm{dt}}{\mathrm{t}^2+1} \\ & =2 \tan ^{-1} \mathrm{t} \\ & =\left.2 \tan ^{-1}\left(\sqrt{\mathrm{e}^{\mathrm{x}}-1}\right)\right|_\alpha ^{\log _{\mathrm{e}}^4} \\ & =2\left[\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}\right]=\frac{\pi}{6} \\ & =\frac{\pi}{3}-\tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}=\frac{\pi}{12} \\ & \Rightarrow \tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}=\frac{\pi}{4} \\ & \mathrm{e}^\alpha=2 \quad \mathrm{e}^{-\alpha}=\frac{1}{2} \\ & \mathrm{x}^2-\left(2+\frac{1}{2}\right) \mathrm{x}+1=0 \\ & 2 \mathrm{x}^2-5 \mathrm{x}+2=0\end{aligned}$$​∫αlog4​ex−1​dx​=6π​ Let ex−1=t2exdx=2tdt=∫t2+12dt​=2tan−1t=2tan−1(ex−1​)​αloge4​​=2[tan−13​−tan−1eα−1​]=6π​=3π​−tan−1eα−1​=12π​⇒tan−1eα−1​=4π​eα=2e−α=21​x2−(2+21​)x+1=02x2−5x+2=0​