Question Number 175191 by ajfour last updated on 22/Aug/22
$$\int\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}{dx}=? \\ $$
Commented by ajfour last updated on 22/Aug/22
$${Thank}\:{you}\:{both}\:{sirs}. \\ $$
Answered by Ar Brandon last updated on 22/Aug/22
$${I}=\int\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}{dx}=\int\sqrt{\frac{{x}−\mathrm{1}}{{x}}}{dx}=\int\frac{{x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −{x}}}{dx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −{x}}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\:\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}}{dx} \\ $$$$\:\:=\sqrt{{x}^{\mathrm{2}} −{x}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcosh}\left(\mathrm{2}{x}−\mathrm{1}\right)+{C} \\ $$$$\:\:=\sqrt{{x}^{\mathrm{2}} −{x}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\left(\mathrm{2}{x}−\mathrm{1}\right)+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}}\mid+{C} \\ $$
Answered by MJS_new last updated on 22/Aug/22
![∫(√(1−(1/x)))dx=∫(√((x−1)/x))dx= [t=(√(x/(x−1))) → dx=−2(x−1)^2 (√(x/(x−1)))dt] =−2∫(dt/((t^2 −1)^2 ))=(t/(t^2 −1))+(1/2)ln ((t−1)/(t+1)) = =(√(x(x−1)))+ln ((√x)−(√(x−1))) +C](https://www.tinkutara.com/question/Q175192.png)
$$\int\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}{dx}=\int\sqrt{\frac{{x}−\mathrm{1}}{{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{{x}}{{x}−\mathrm{1}}}\:\rightarrow\:{dx}=−\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{\frac{{x}}{{x}−\mathrm{1}}}{dt}\right] \\ $$$$=−\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:= \\ $$$$=\sqrt{{x}\left({x}−\mathrm{1}\right)}+\mathrm{ln}\:\left(\sqrt{{x}}−\sqrt{{x}−\mathrm{1}}\right)\:+{C} \\ $$