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Question Number 206096 by RoseAli last updated on 06/Apr/24

$$\int \frac{1}{x^3\sqrt{x^2-1}}.dx$$

Answered by Frix last updated on 07/Apr/24

$$\int \frac{dx}{x^3\sqrt{x^2-1}}\stackrel{t=\sqrt{x^2-1}}{=}\int \frac{dt}{{(t^2+1)}^2}$$$$\mathrm{From}\mathrm{here}\mathrm{we}\mathrm{could}\mathrm{use}{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}$$$$\mathrm{Method}\mathrm{or}\mathrm{this}:$$$$\int \frac{dt}{{(t^2+1)}^2}\stackrel{u={\tan }^{-1}t}{=}\int {\cos }^{2}udu=$$$$=\frac{1}{2}\int (1+\cos 2u)du=$$$$=\frac{u}{2}+\frac{\sin 2u}{4}=\frac{{\tan }^{-1}t}{2}+\frac{t}{2(t^2+1)}=$$$$=\frac{{\tan }^{-1}\sqrt{x^2-1}}{2}+\frac{\sqrt{x^2-1}}{2x}+C$$

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