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Question Number 131288 by mohammad17 last updated on 03/Feb/21

Answered by liberty last updated on 11/Feb/21

$$\mathrm{L}=\int \sqrt{\frac{1-{3\mathrm{x}}^{-3}}{{\mathrm{x}}^8}}\mathrm{dx}=\int {\mathrm{x}}^{-4}\sqrt{1-{3\mathrm{x}}^{-3}}\mathrm{dx}$$$$\mathrm{change}\mathrm{of}\mathrm{variable}:\sqrt{1-{3\mathrm{x}}^{-3}}=\mathrm{h}\mathrm{or}1-{3\mathrm{x}}^{-3}={\mathrm{h}}^2$$$$\Rightarrow {9\mathrm{x}}^{-4}\mathrm{dx}=2\mathrm{h}\mathrm{dh}$$$$\mathrm{L}=\frac{2}{9}\int {\mathrm{h}}^2\mathrm{dh}=\frac{2}{27}{\mathrm{h}}^3+\mathrm{c}$$$$\mathrm{L}=\frac{2}{27}\sqrt{{\left(\frac{{\mathrm{x}}^3-3}{{\mathrm{x}}^3}\right)}^3}+\mathrm{c}=\frac{2}{27}\left(\frac{{\mathrm{x}}^3-3}{{\mathrm{x}}^3}\right)\sqrt{\frac{{\mathrm{x}}^3-3}{{\mathrm{x}}^3}}+\mathrm{c}$$

Answered by Dwaipayan Shikari last updated on 03/Feb/21

$$\int \sqrt{\frac{x^3-3}{x^{11}}}dx$$$$=\int \frac{1}{x^4}\sqrt{1-\frac{3}{x^3}}dx$$$$=\frac{1}{9}\int \frac{9}{x^4}\sqrt{1-\frac{3}{x^3}}dx=\frac{2}{27}{(1-\frac{3}{x^3})}^{\frac{3}{2}}+C$$

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