Question Number 212048 by Spillover last updated on 28/Sep/24
Answered by Ghisom last updated on 28/Sep/24
![∫_0 ^∞ (dx/( (√x)(√(x^2 +1))(√(x+(√(x^2 +1))))))= [t=x+(√(x^2 +1))] =(√2)∫_1 ^∞ (dt/(t(√(t^2 −1))))= [u=t+(√(t^2 −1))] =2(√2)∫_1 ^∞ (du/(u^2 +1))=[2(√2)arctan u]_1 ^∞ = =(((√2)π)/2)](https://www.tinkutara.com/question/Q212062.png)
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\sqrt{\mathrm{2}}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dt}}{{t}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{u}={t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right] \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\left[\mathrm{2}\sqrt{\mathrm{2}}\mathrm{arctan}\:{u}\right]_{\mathrm{1}} ^{\infty} = \\ $$$$=\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{2}} \\ $$
Commented by Spillover last updated on 28/Sep/24
$${great}.{thanks} \\ $$