Question Number 168473 by Mastermind last updated on 11/Apr/22
$${Integrate}: \\ $$$$\int\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by MJS_new last updated on 11/Apr/22
![∫(√((1−x)/x))dx= [t=(√((1−x)/x)) → dx=−2x^2 tdt] =−2∫(t^2 /((t^2 +1)^2 ))dt=(t/(t^2 +1))−arctan t = =(√((1−x)x))−arctan (√((1−x)/x)) +C](https://www.tinkutara.com/question/Q168484.png)
$$\int\sqrt{\frac{\mathrm{1}−{x}}{{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{\mathrm{1}−{x}}{{x}}}\:\rightarrow\:{dx}=−\mathrm{2}{x}^{\mathrm{2}} {tdt}\right] \\ $$$$=−\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}=\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}−\mathrm{arctan}\:{t}\:= \\ $$$$=\sqrt{\left(\mathrm{1}−{x}\right){x}}−\mathrm{arctan}\:\sqrt{\frac{\mathrm{1}−{x}}{{x}}}\:+{C} \\ $$
Commented by Mastermind last updated on 12/Apr/22
$$ \\ $$$${Thanks}\:{man} \\ $$