Question Number 47062 by maxmathsup by imad last updated on 04/Nov/18
$${find}\:\int\:\:\:\sqrt{\frac{\sqrt{{x}}−\mathrm{1}}{\:\sqrt{{x}}+\mathrm{1}}}{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18
![t^2 =x dx=2tdt ∫(√((t−1)/(t+1))) ×2tdt ∫((2t(t−1))/( (√(t^2 −1))))dt ∫((2t^2 −2+2−2t)/( (√(t^2 −1))))dt 2∫(√(t^2 −1)) +2∫(dt/( (√(t^2 −1))))−∫((d(t^2 −1))/( (√(t^2 −1)))) 2[(t/2)(√(t^2 −1)) −(1^2 /2)ln∣t+(√(t^2 −1)) ]+2ln∣t+(√(t^2 −1)) ∣−(((t^2 −1)^(1/2) )/(1/2))+c =2[((√x)/2)(√(x−1)) −(1/2)ln∣(√x) +(√(x−1)) ∣]+2ln∣(√x) +(√(x−))1 ∣−(((x−1)^(1/2) )/(1/2))+c](https://www.tinkutara.com/question/Q47089.png)
$${t}^{\mathrm{2}} ={x}\:\:\:{dx}=\mathrm{2}{tdt} \\ $$$$\int\sqrt{\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}}\:×\mathrm{2}{tdt} \\ $$$$\int\frac{\mathrm{2}{t}\left({t}−\mathrm{1}\right)}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{dt} \\ $$$$\int\frac{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}+\mathrm{2}−\mathrm{2}{t}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{dt} \\ $$$$\mathrm{2}\int\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:+\mathrm{2}\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}−\int\frac{{d}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\mathrm{2}\left[\frac{{t}}{\mathrm{2}}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}{ln}\mid{t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:\right]+\mathrm{2}{ln}\mid{t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:\mid−\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}+{c} \\ $$$$=\mathrm{2}\left[\frac{\sqrt{{x}}}{\mathrm{2}}\sqrt{{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\sqrt{{x}}\:+\sqrt{{x}−\mathrm{1}}\:\mid\right]+\mathrm{2}{ln}\mid\sqrt{{x}}\:+\sqrt{{x}−}\mathrm{1}\:\mid−\frac{\left({x}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}+{c} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 05/Nov/18
$${thank}\:{you}\:{sir}\:{Tanmay}. \\ $$