Let _ ^ _e4dxe^x-1=( )/(6).

Question

Let $\int _{\alpha }^{{\log }_{e}4}\frac{dx}{\sqrt{{e}^{x}-1}}=\frac{\pi }{6}$. Then ${e}^{\alpha }$ and ${e}^{-\alpha }$ are the roots of the equation :

Accepted Answer

$\begin{gathered} \sqrt{{e}^{x}-1}=t \ {e}^{x}-1={t}^{2} \ {e}^{x}=1+{t}^{2} \ {e}^{x}=0+2t-\frac{dt}{dx} \ \frac{dt}{dx}=\frac{{e}^{x}}{2t}=\frac{{t}^{2}+1}{2t} \ I=\int \frac{2t}{t(1+{t}^{2})}dt=2{	an }^{-1}t \ \Rightarrow \, I=\int _{\alpha }^{{\log }^{4}}\frac{dx}{\sqrt{{e}^{x}-1}} \ I={2{	an }^{-1}\sqrt{{e}^{x}-1}|}_{\log 4}^{\alpha } \ =2({	an }^{-1}\sqrt{3}-{	an }^{-1}\sqrt{{e}^{\alpha }-1})=\frac{\pi }{6} \ \Rightarrow \frac{\pi }{3}-{	an }^{-1}(\sqrt{{e}^{\alpha }-1})=\frac{\pi }{12} \ {	an }^{-1}(\sqrt{{e}^{\alpha }-1})=\frac{\pi }{4} \ \Rightarrow {e}^{\alpha }-1=1 \ {e}^{\alpha }=2\Rightarrow {e}^{-\alpha }=\frac{1}{2} \end{gathered}$

$	herefore \,$ Quadratic equation whose roots are ${e}^{a}$ \& ${e}^{-\alpha }$ is

$\begin{gathered} {x}^{2}-({e}^{\alpha }+{e}^{-\alpha })x+{e}^{\alpha }	imes {e}^{-\alpha }=0 \ {x}^{2}-(2+\frac{1}{2})x+1=0 \ 2{x}^{2}-5x+2=0 \end{gathered}$

Solutions · 1