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Question Number 104388 by Ar Brandon last updated on 21/Jul/20

$$\int \frac{\mathrm{d}x}{x^2\sqrt[3]{x^3-1}}$$

Answered by Dwaipayan Shikari last updated on 21/Jul/20

$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^3{(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{1}{3}}}=\int \frac{\frac{1}{{\mathrm{x}}^4}.\mathrm{x}}{{(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{1}{3}}}\mathrm{dx}=\frac{\mathrm{x}}{3}\int \frac{\frac{3}{{\mathrm{x}}^4}}{{(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{1}{3}}}-\frac{1}{3}\int \int \frac{\frac{3}{{\mathrm{x}}^4}}{{(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{1}{3}}}$$$$=\frac{\mathrm{x}}{2}{(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{2}{3}}-\frac{1}{2}\int {(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{2}{3}}$$$$=\frac{\mathrm{x}}{2}{(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{2}{3}}-\frac{1}{2}\int \frac{3}{{\mathrm{x}}^4}.\frac{{\mathrm{x}}^4}{3}{(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{2}{3}}\mathrm{dx}$$$$\frac{\mathrm{x}}{2}{(1-\frac{1}{{\mathrm{x}}^3})}^{\frac{2}{3}}-\frac{1}{6}\int {\mathrm{x}}^4{\mathrm{t}}^{\frac{2}{3}}\mathrm{dt}\{1-\frac{1}{{\mathrm{x}}^3}=\mathrm{t}{\mathrm{x}}^3=\frac{1}{1-\mathrm{t}}$$$$\frac{\mathrm{x}}{2}(1-\frac{1}{{\mathrm{x}}^3})-\frac{1}{6}\int {\left(\frac{1}{1-\mathrm{t}}\right)}^{\frac{4}{3}}{\mathrm{t}}^{\frac{2}{3}}.....\mathrm{continue}$$$$$$$$$$

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