Question Number 158965 by cortano last updated on 11/Nov/21
$$\:\int\:\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{{x}}\:+\mathrm{1}}\:{dx}\:=? \\ $$
Answered by puissant last updated on 11/Nov/21
$$\Omega=\int\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{{x}}+\mathrm{1}}{dx} \\ $$$${u}=\sqrt{{x}}\:\rightarrow\:{du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}}}\:\rightarrow\:{dx}=\mathrm{2}{udu} \\ $$$$\Rightarrow\:\Omega\:=\:\int\frac{\mathrm{2}{u}\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}{{u}+\mathrm{1}}{du}\:;\: \\ $$$${u}={tan}\left({t}\right)\:\rightarrow\:{du}=\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({t}\right)\right){dt} \\ $$$$\Omega\:=\:\mathrm{2}\int\frac{{tan}\left({t}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({t}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{tan}\left({t}\right)+\mathrm{1}}{dt} \\ $$$$………………… \\ $$
Answered by Ar Brandon last updated on 11/Nov/21
![I=∫((√(1+x))/( (√x)+1))dx, x=t^2 ⇒dx=2tdt =2∫((√(1+t^2 ))/(t+1))tdt=2∫((√(1+t^2 ))−((√(1+t^2 ))/(1+t)))dt ∫(√(1+t^2 ))dt, t=sinhϑ⇒dt=coshϑdϑ =(1/2)∫(cosh2ϑ+1)dϑ=(1/2)[((sinh2ϑ)/2)+ϑ] ∫((√(1+t^2 ))/(1+t))dt=∫((1+t^2 )/((1+t)(√(1+t^2 ))))=∫(((t−1)/( (√(1+t^2 ))))+(2/((t+1)(√(1+t^2 )))))dt =(√(1+t^2 ))−argsh(t)+∫(2/((t+1)(√(1+t^2 ))))dt (1/(t+1))=u⇒t=(1/u)−1⇒dt=−(1/u^2 )du ∫(2/((t+1)(√(1+t^2 ))))dt=∫(2/((1/u)(√(1+((1/u)−1)^2 ))))∙(−(du/u^2 )) =−∫((2du)/( (√(u^2 +1−2u+u^2 ))))=−∫((2du)/( (√(2u^2 −2u+1))))=−(√2)∫(du/( (√((u−(1/2))^2 +(1/4))))) =−(√2)argsh(2u−1)+C](https://www.tinkutara.com/question/Q158975.png)
$${I}=\int\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{{x}}+\mathrm{1}}{dx},\:{x}={t}^{\mathrm{2}} \Rightarrow{dx}=\mathrm{2}{tdt} \\ $$$$\:\:\:=\mathrm{2}\int\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}+\mathrm{1}}{tdt}=\mathrm{2}\int\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }−\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+{t}}\right){dt} \\ $$$$\int\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt},\:{t}=\mathrm{sinh}\vartheta\Rightarrow{dt}=\mathrm{cosh}\vartheta{d}\vartheta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{cosh2}\vartheta+\mathrm{1}\right){d}\vartheta=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{sinh2}\vartheta}{\mathrm{2}}+\vartheta\right] \\ $$$$\int\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+{t}}{dt}=\int\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}\right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}=\int\left(\frac{{t}−\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}+\frac{\mathrm{2}}{\left({t}+\mathrm{1}\right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{argsh}\left({t}\right)+\int\frac{\mathrm{2}}{\left({t}+\mathrm{1}\right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt} \\ $$$$\frac{\mathrm{1}}{{t}+\mathrm{1}}={u}\Rightarrow{t}=\frac{\mathrm{1}}{{u}}−\mathrm{1}\Rightarrow{dt}=−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }{du} \\ $$$$\int\frac{\mathrm{2}}{\left({t}+\mathrm{1}\right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt}=\int\frac{\mathrm{2}}{\frac{\mathrm{1}}{{u}}\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}}{{u}}−\mathrm{1}\right)^{\mathrm{2}} }}\centerdot\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right) \\ $$$$=−\int\frac{\mathrm{2}{du}}{\:\sqrt{{u}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{u}+{u}^{\mathrm{2}} }}=−\int\frac{\mathrm{2}{du}}{\:\sqrt{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}{u}+\mathrm{1}}}=−\sqrt{\mathrm{2}}\int\frac{{du}}{\:\sqrt{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$$$=−\sqrt{\mathrm{2}}\mathrm{argsh}\left(\mathrm{2}{u}−\mathrm{1}\right)+{C} \\ $$