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1-x-3-x-2-1-dx-




Question Number 206096 by RoseAli last updated on 06/Apr/24
∫(1/(x^3 (√(x^2 −1)))) .dx
$$\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:.{dx} \\ $$
Answered by Frix last updated on 07/Apr/24
∫(dx/(x^3 (√(x^2 −1)))) =^(t=(√(x^2 −1)))  ∫(dt/((t^2 +1)^2 ))  From here we could use Ostrogradski′s  Method or this:  ∫(dt/((t^2 +1)^2 )) =^(u=tan^(−1)  t)  ∫cos^2  u du=  =(1/2)∫(1+cos 2u)du=  =(u/2)+((sin 2u)/4)=((tan^(−1)  t)/2)+(t/(2(t^2 +1)))=  =((tan^(−1)  (√(x^2 −1)))/2)+((√(x^2 −1))/(2x))+C
$$\int\frac{{dx}}{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\overset{{t}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} {=}\:\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{From}\:\mathrm{here}\:\mathrm{we}\:\mathrm{could}\:\mathrm{use}\:\mathrm{Ostrogradski}'\mathrm{s} \\ $$$$\mathrm{Method}\:\mathrm{or}\:\mathrm{this}: \\ $$$$\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\overset{{u}=\mathrm{tan}^{−\mathrm{1}} \:{t}} {=}\:\int\mathrm{cos}^{\mathrm{2}} \:{u}\:{du}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{u}\right){du}= \\ $$$$=\frac{{u}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}{u}}{\mathrm{4}}=\frac{\mathrm{tan}^{−\mathrm{1}} \:{t}}{\mathrm{2}}+\frac{{t}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{tan}^{−\mathrm{1}} \:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}}+\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}{x}}+{C} \\ $$

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