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Question Number 127851 by bemath last updated on 02/Jan/21

$$\psi =\int \frac{\mathrm{dx}}{{\mathrm{x}}^3\sqrt[5]{{({\mathrm{x}}^5+1)}^3}}?$$

Answered by liberty last updated on 02/Jan/21

$$\psi =\int \frac{\mathrm{dx}}{{\mathrm{x}}^3\sqrt[5]{{\mathrm{x}}^{15}{(1+{\mathrm{x}}^{-5})}^3}}$$$$\psi =\int \frac{\mathrm{dx}}{{\mathrm{x}}^6\sqrt[5]{{(1+{\mathrm{x}}^{-5})}^3}}=\int \frac{{\mathrm{x}}^{-6}\mathrm{dx}}{\sqrt[5]{{(1+{\mathrm{x}}^{-5})}^3}}$$$$\mathrm{let}\mathrm{t}=1+{\mathrm{x}}^{-5}\Rightarrow \mathrm{dt}=-{5\mathrm{x}}^{-6}\mathrm{dx}$$$$\psi =-\frac{1}{5}\int \frac{\mathrm{dt}}{{\mathrm{t}}^{3/5}}=-\frac{1}{5}\int {\mathrm{t}}^{-3/5}\mathrm{dt}$$$$\psi =-\frac{1}{5}.\frac{5}{2}{\mathrm{t}}^{2/5}+\mathrm{C}=-\frac{\sqrt[5]{{({\mathrm{x}}^5+1)}^2}}{{2\mathrm{x}}^2}+\mathrm{C}$$

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