9514 1404 393
Answer:
4arcsin(x^(1/4)) +C
Step-by-step explanation:
Let x = z^4. Then dx = 4z^3·dz, and the given expression is ...
[tex]\displaystyle\frac{dx}{\sqrt{x\sqrt{x}-x^2}}=\frac{4z^3\,dz}{\sqrt{z^4z^2-z^8}}=4\cdot\frac{dz}{\sqrt{1-z^2}}[/tex]
and its integral is ...
[tex]\displaystyle\int{\frac{4\,dz}{\sqrt{1-z^2}}}=4\arcsin{(z)}+C=\boxed{4\arcsin{\sqrt[4]{x}}+C}[/tex]
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Note that the original integrand is only defined on the interval (0, 1).
