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

x \to 1 lim (\frac{17 - x ^{2} - 4}{x - 1})

\frac{x ^{2}}{18 x - 9}

\frac{d}{d x} (\frac{4}{5} x^{15})

(10.9) (40.3)

\int sin (2 π) x d x

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