Question Number 168857 by Sotoberry last updated on 19/Apr/22
Answered by MJS_new last updated on 20/Apr/22
![∫(1/x^3 )(√((x^2 −1)/(x+1)))dx=∫((√(x−1))/x^3 )dx= [t=(√(x−1)) → dx=2(√(x−1))dt] =2∫(t^2 /((t^2 +1)^3 ))dt=((t(t^2 −1))/(4(t^2 +1)^2 ))+(1/4)arctan t = =(((x−2)(√(x−1)))/(4x^2 ))+(1/4)arctan (√(x−1)) +C](https://www.tinkutara.com/question/Q168870.png)
$$\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\sqrt{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}}{dx}=\int\frac{\sqrt{{x}−\mathrm{1}}}{{x}^{\mathrm{3}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}−\mathrm{1}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}−\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dt}=\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arctan}\:{t}\:= \\ $$$$=\frac{\left({x}−\mathrm{2}\right)\sqrt{{x}−\mathrm{1}}}{\mathrm{4}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arctan}\:\sqrt{{x}−\mathrm{1}}\:+{C} \\ $$
Answered by MJS_new last updated on 20/Apr/22
![∫(√((x−1)/x^5 ))dx= t=(√((x−1)/x)) → dx=2x^2 (√((x−1)/x))dt] =2∫t^2 dt=((2t^3 )/3)=(2/3)(√(((x−1)^3 )/x^3 ))+C](https://www.tinkutara.com/question/Q168871.png)
$$\int\sqrt{\frac{{x}−\mathrm{1}}{{x}^{\mathrm{5}} }}{dx}= \\ $$$$\left.\:\:\:\:\:{t}=\sqrt{\frac{{x}−\mathrm{1}}{{x}}}\:\rightarrow\:{dx}=\mathrm{2}{x}^{\mathrm{2}} \sqrt{\frac{{x}−\mathrm{1}}{{x}}}{dt}\right] \\ $$$$=\mathrm{2}\int{t}^{\mathrm{2}} {dt}=\frac{\mathrm{2}{t}^{\mathrm{3}} }{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{{x}^{\mathrm{3}} }}+{C} \\ $$
Answered by cortano1 last updated on 20/Apr/22
![∫ ((√(x−1))/( (√x^5 ))) dx= ∫ (((√(1−x^(−1) )) dx)/x^2 ) = ∫ x^(−2) (√(1−x^(−1) )) dx [ let u^2 =1−x^(−1) ⇒2u du = x^(−2) dx ] I= ∫ u (2u du)= (2/3)u^3 +c = (2/3) (((x−1)/x))(√((x−1)/x)) + c](https://www.tinkutara.com/question/Q168893.png)
$$\:\int\:\frac{\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}^{\mathrm{5}} }}\:{dx}=\:\int\:\frac{\sqrt{\mathrm{1}−{x}^{−\mathrm{1}} }\:{dx}}{{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int\:{x}^{−\mathrm{2}} \:\sqrt{\mathrm{1}−{x}^{−\mathrm{1}} }\:{dx} \\ $$$$\:\left[\:{let}\:{u}^{\mathrm{2}} =\mathrm{1}−{x}^{−\mathrm{1}} \:\Rightarrow\mathrm{2}{u}\:{du}\:=\:{x}^{−\mathrm{2}} \:{dx}\:\right] \\ $$$$\:{I}=\:\int\:{u}\:\left(\mathrm{2}{u}\:{du}\right)=\:\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}} +{c} \\ $$$$\:\:\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)\sqrt{\frac{{x}−\mathrm{1}}{{x}}}\:+\:{c} \\ $$$$ \\ $$