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Question Number 159938 by ZiYangLee last updated on 22/Nov/21

$$\mathrm{Evaluate}{\int }_1^4\sqrt{\frac{x-1}{x^5}}dx.$$

Answered by Ar Brandon last updated on 22/Nov/21

$$I={\int }_1^4\sqrt{\frac{x-1}{x^5}}dx={\int }_1^4\frac{1}{x^2}\sqrt{1-\frac{1}{x}}dx$$$$u=1-\frac{1}{x}\Rightarrow du=\frac{1}{x^2}dx$$$$I={\int }_0^{\frac{3}{4}}\sqrt{u}du=\frac{2}{3}{\left[u^{\frac{3}{2}}\right]}_0^{\frac{3}{4}}=\frac{2}{3}{\left(\frac{\sqrt{3}}{2}\right)}^3=\frac{\sqrt{3}}{4}$$

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