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Question Number 144234 by mathdanisur last updated on 23/Jun/21

$$\int \frac{\sqrt{x^2-x}}{x^3}dx=?$$

Answered by liberty last updated on 23/Jun/21

$$\int \frac{\sqrt{1-{\mathrm{x}}^{-1}}}{{\mathrm{x}}^2}\mathrm{dx}=\int {\mathrm{x}}^{-2}\sqrt{1-{\mathrm{x}}^{-1}}\mathrm{dx}$$$$\mathrm{let}{\mathrm{u}}^2=1-{\mathrm{x}}^{-1}\to 2\mathrm{u}\mathrm{du}={\mathrm{x}}^{-2}\mathrm{dx}$$$$\mathrm{then}\int \sqrt{{\mathrm{u}}^2}\left(2\mathrm{u}\mathrm{du}\right)=\int {2\mathrm{u}}^2\mathrm{du}$$$$=\frac{2}{3}{\mathrm{u}}^3+\mathrm{c}=\frac{2}{3}\sqrt{{(1-{\mathrm{x}}^{-1})}^3}+\mathrm{c}$$$$=\frac{2\sqrt{{(\mathrm{x}-1)}^3}}{3\mathrm{x}\sqrt{\mathrm{x}}}+\mathrm{c}$$$$=\frac{2(\mathrm{x}-1)\sqrt{\mathrm{x}-1}}{3\mathrm{x}\sqrt{\mathrm{x}}}+\mathrm{c}$$

Commented by mathdanisur last updated on 23/Jun/21

$$ThankyouSir$$

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