Question Number 151761 by talminator2856791 last updated on 22/Aug/21
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{1}}{\:\sqrt{{e}^{{x}} +\mathrm{1}}}\:{dx}\: \\ $$$$\: \\ $$
Answered by mindispower last updated on 22/Aug/21
![∫_0 ^∞ ((e^(−(x/2)) dx)/( (√(1+(e^(−(x/2)) )^2 ))))=−2∫_0 ^∞ (de^(−(x/2)) /( (√(1+(e^(−(x/2)) )^2 )))) =−2[argsh(e^(−(x/2)) )]_0 ^∞ =2argsh(1)=ln(3+2(√2))](https://www.tinkutara.com/question/Q151764.png)
$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\frac{{x}}{\mathrm{2}}} {dx}}{\:\sqrt{\mathrm{1}+\left({e}^{−\frac{{x}}{\mathrm{2}}} \right)^{\mathrm{2}} }}=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{de}^{−\frac{{x}}{\mathrm{2}}} }{\:\sqrt{\mathrm{1}+\left({e}^{−\frac{{x}}{\mathrm{2}}} \right)^{\mathrm{2}} }} \\ $$$$=−\mathrm{2}\left[{argsh}\left({e}^{−\frac{{x}}{\mathrm{2}}} \right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\mathrm{2}{argsh}\left(\mathrm{1}\right)={ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$