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Question Number 113760 by gloriousman last updated on 15/Sep/20

$$\int \sqrt{\frac{\mathrm{x}-1}{{\mathrm{x}}^5}}\mathrm{dx}$$

Answered by bemath last updated on 15/Sep/20

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

Commented by gloriousman last updated on 15/Sep/20

$$\mathrm{Thanks}$$

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