Question Number 113760 by gloriousman last updated on 15/Sep/20
$$\int\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }}\mathrm{dx} \\ $$
Answered by bemath last updated on 15/Sep/20
![I=∫ (1/x^2 )(√((x−1)/x)) dx I= ∫ (1/x^2 ) (√(1−(1/x))) dx [ let u = 1−(1/x) ⇒du = (dx/x^2 ) ] I= ∫ u^(1/2) du = (2/3)u^(3/2) + c I= (2/3) (((x−1)/x))(√((x−1)/x)) + c](https://www.tinkutara.com/question/Q113779.png)
$${I}=\int\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\sqrt{\frac{{x}−\mathrm{1}}{{x}}}\:{dx}\: \\ $$$${I}=\:\int\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:{dx}\: \\ $$$$\left[\:{let}\:{u}\:=\:\mathrm{1}−\frac{\mathrm{1}}{{x}}\:\Rightarrow{du}\:=\:\frac{{dx}}{{x}^{\mathrm{2}} }\:\right] \\ $$$${I}=\:\int\:{u}^{\frac{\mathrm{1}}{\mathrm{2}}} \:{du}\:=\:\frac{\mathrm{2}}{\mathrm{3}}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} +\:{c} \\ $$$${I}=\:\frac{\mathrm{2}}{\mathrm{3}}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)\sqrt{\frac{{x}−\mathrm{1}}{{x}}}\:+\:{c} \\ $$
Commented by gloriousman last updated on 15/Sep/20
$$\mathrm{Thanks} \\ $$