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Question Number 168857 by Sotoberry last updated on 19/Apr/22

Answered by MJS_new last updated on 20/Apr/22

$$\int \frac{1}{x^3}\sqrt{\frac{x^2-1}{x+1}}dx=\int \frac{\sqrt{x-1}}{x^3}dx=$$$$[t=\sqrt{x-1}\to dx=2\sqrt{x-1}dt]$$$$=2\int \frac{t^2}{{(t^2+1)}^3}dt=\frac{t(t^2-1)}{4{(t^2+1)}^2}+\frac{1}{4}\arctan t=$$$$=\frac{(x-2)\sqrt{x-1}}{4x^2}+\frac{1}{4}\arctan \sqrt{x-1}+C$$

Answered by MJS_new last updated on 20/Apr/22

$$\int \sqrt{\frac{x-1}{x^5}}dx=$$$$t=\sqrt{\frac{x-1}{x}}\to dx=2x^2\sqrt{\frac{x-1}{x}}dt]$$$$=2\int t^{2}dt=\frac{2t^3}{3}=\frac{2}{3}\sqrt{\frac{{(x-1)}^3}{x^3}}+C$$

Answered by cortano1 last updated on 20/Apr/22

$$\int \frac{\sqrt{x-1}}{\sqrt{x^5}}dx=\int \frac{\sqrt{1-x^{-1}}dx}{x^2}$$$$=\int x^{-2}\sqrt{1-x^{-1}}dx$$$$[let{u}^2=1-x^{-1}\Rightarrow 2udu=x^{-2}dx]$$$$I=\int u\left(2udu\right)=\frac{2}{3}{u}^3+c$$$$=\frac{2}{3}\left(\frac{x-1}{x}\right)\sqrt{\frac{x-1}{x}}+c$$$$$$

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