Question Number 53697 by gunawan last updated on 25/Jan/19
$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\underset{\mathrm{log}\:\mathrm{2}} {\overset{{x}} {\int}}\:\:\frac{\mathrm{1}}{\:\sqrt{{e}^{{x}} −\mathrm{1}}}\:{dx}=\:\frac{\pi}{\mathrm{6}}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$
Commented by Abdo msup. last updated on 25/Jan/19
![let A(x)=∫_(log2) ^x (dt/( (√(e^t −1))))⇒A(x)=_(e^t =u) ∫_2 ^e^x (du/(u(√(u−1)))) =_((√(u−1))=α) ∫_1 ^(√(e^x −1)) ((2αdα)/((1+α^2 )α)) =2 ∫_1 ^(√(e^x −1)) (dα/(1+α^2 )) =2[arctan(α)]_1 ^(√(e^x −1)) =2{ arctan((√(e^x −1)))−(π/4)} =2arctan((√(e^x −1)))−(π/2) so A(x)=(π/6) ⇔2 arctan((√(e^x −1)))=(π/2) +(π/6) ⇒ 2 arctan((√(e^x −1)))=((4π)/6) =((2π)/3) ⇒ artan((√(e^x −1)))=(π/3) ⇒(√(e^x −1))=tan((π/3)) ⇒ (√(e^x −1))=(√3) ⇒e^x −1 =3 ⇒e^x =4 ⇒x =2ln(2) .](https://www.tinkutara.com/question/Q53707.png)
$${let}\:{A}\left({x}\right)=\int_{{log}\mathrm{2}} ^{{x}} \:\frac{{dt}}{\:\sqrt{{e}^{{t}} −\mathrm{1}}}\Rightarrow{A}\left({x}\right)=_{{e}^{{t}} ={u}} \:\:\:\int_{\mathrm{2}} ^{{e}^{{x}} } \:\:\:\frac{{du}}{{u}\sqrt{{u}−\mathrm{1}}} \\ $$$$=_{\sqrt{{u}−\mathrm{1}}=\alpha} \:\:\:\:\:\int_{\mathrm{1}} ^{\sqrt{{e}^{{x}} −\mathrm{1}}} \:\:\:\frac{\mathrm{2}\alpha{d}\alpha}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\alpha}\:=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{{e}^{{x}} −\mathrm{1}}} \:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$$$=\mathrm{2}\left[{arctan}\left(\alpha\right)\right]_{\mathrm{1}} ^{\sqrt{{e}^{{x}} −\mathrm{1}}} =\mathrm{2}\left\{\:{arctan}\left(\sqrt{{e}^{{x}} −\mathrm{1}}\right)−\frac{\pi}{\mathrm{4}}\right\} \\ $$$$=\mathrm{2}{arctan}\left(\sqrt{{e}^{{x}} −\mathrm{1}}\right)−\frac{\pi}{\mathrm{2}}\:{so} \\ $$$${A}\left({x}\right)=\frac{\pi}{\mathrm{6}}\:\:\Leftrightarrow\mathrm{2}\:{arctan}\left(\sqrt{{e}^{{x}} −\mathrm{1}}\right)=\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{6}}\:\Rightarrow \\ $$$$\mathrm{2}\:{arctan}\left(\sqrt{{e}^{{x}} −\mathrm{1}}\right)=\frac{\mathrm{4}\pi}{\mathrm{6}}\:=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\Rightarrow \\ $$$${artan}\left(\sqrt{{e}^{{x}} −\mathrm{1}}\right)=\frac{\pi}{\mathrm{3}}\:\Rightarrow\sqrt{{e}^{{x}} −\mathrm{1}}={tan}\left(\frac{\pi}{\mathrm{3}}\right)\:\Rightarrow \\ $$$$\sqrt{{e}^{{x}} −\mathrm{1}}=\sqrt{\mathrm{3}}\:\Rightarrow{e}^{{x}} −\mathrm{1}\:=\mathrm{3}\:\Rightarrow{e}^{{x}} =\mathrm{4}\:\Rightarrow{x}\:=\mathrm{2}{ln}\left(\mathrm{2}\right)\:. \\ $$