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Question Number 108776 by bobhans last updated on 19/Aug/20

$$\stackrel{bobhans}{⋮}$$$$\int \frac{dx}{\sqrt{x\sqrt{x}-x^2}}=?$$

Answered by john santu last updated on 19/Aug/20

$$\frac{⊸\mathcal{JS}⊸}{\mathrm{♡}}$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^{\frac{3}{2}}-{\mathrm{x}}^2}}=\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^{\frac{3}{2}}(1-{\mathrm{x}}^{\frac{1}{2}})}}$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^{\frac{3}{4}}\sqrt{1-{\mathrm{x}}^{\frac{1}{2}}}}.[\mathrm{let}{\mathrm{x}}^{\frac{1}{4}}=\mathrm{z}]$$$$\mathrm{I}=\int \frac{{4\mathrm{z}}^3\mathrm{dz}}{{\mathrm{z}}^3\sqrt{1-{\mathrm{z}}^2}}=4\int \frac{\mathrm{dz}}{\sqrt{1-{\mathrm{z}}^2}}$$$$\mathrm{I}=4\mathrm{arc}\sin \left(\mathrm{z}\right)+\mathrm{c}$$$$\mathrm{I}=4\mathrm{arc}\sin \left(\sqrt[4]{\mathrm{x}}\right)+\mathrm{c}$$

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