What is the integral of 1/(x^2*\sqrt[5]{x^2)} ?

The integral of 1/(x^2*\sqrt[5]{x^2)} is -5/(7x^{7/5)}+C

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integral of 1/(x^2*\sqrt[5]{x^2)}

- \frac{5}{7 x ^{\frac{7}{5}}} + C

Solution steps

assuming

x \geq 0

= \int \frac{1}{x ^{2} x ^{\frac{2}{5}}} d x

Simplify

\frac{1}{x ^{2} x ^{\frac{2}{5}}} : \frac{1}{x ^{\frac{12}{5}}}

= \int \frac{1}{x ^{\frac{12}{5}}} d x

Apply the Power Rule

: - \frac{5}{7 x ^{\frac{7}{5}}}

= - \frac{5}{7 x ^{\frac{7}{5}}}

Add a constant to the solution

= - \frac{5}{7 x ^{\frac{7}{5}}} + C

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What is the integral of 1/(x^2*\sqrt[5]{x^2)} ?
The integral of 1/(x^2*\sqrt[5]{x^2)} is -5/(7x^{7/5)}+C