What is the integral of (\sqrt[3]{x^2})/(x^7) ?

The integral of (\sqrt[3]{x^2})/(x^7) is -3/(16x^{16/3)}+C

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integral of (\sqrt[3]{x^2})/(x^7)

- \frac{3}{16 x ^{\frac{16}{3}}} + C

Solution steps

assuming

x \geq 0

= \int \frac{x ^{\frac{2}{3}}}{x ^{7}} d x

Simplify

\frac{x ^{\frac{2}{3}}}{x ^{7}} : \frac{1}{x ^{\frac{19}{3}}}

= \int \frac{1}{x ^{\frac{19}{3}}} d x

Apply the Power Rule

: - \frac{3}{16 x ^{\frac{16}{3}}}

= - \frac{3}{16 x ^{\frac{16}{3}}}

Add a constant to the solution

= - \frac{3}{16 x ^{\frac{16}{3}}} + C

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What is the integral of (\sqrt[3]{x^2})/(x^7) ?
The integral of (\sqrt[3]{x^2})/(x^7) is -3/(16x^{16/3)}+C