What is the integral of 1/(6\sqrt[6]{x^5)} ?

The integral of 1/(6\sqrt[6]{x^5)} is \sqrt[6]{x}+C

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integral of 1/(6\sqrt[6]{x^5)}

\int \frac{1}{6 6 x ^{5}} d x

6 x + C

Solution steps

\int \frac{1}{6 6 x ^{5}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{1}{6 x ^{5}} d x

6 x^{5} = x^{\frac{5}{6}},

assuming

x \geq 0

= \frac{1}{6} \cdot \int \frac{1}{x ^{\frac{5}{6}}} d x

Apply the Power Rule

= \frac{1}{6} \cdot 6 x^{\frac{1}{6}}

Simplify

\frac{1}{6} \cdot 6 x^{\frac{1}{6}} : 6 x

= 6 x

Add a constant to the solution

= 6 x + C

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What is the integral of 1/(6\sqrt[6]{x^5)} ?
The integral of 1/(6\sqrt[6]{x^5)} is \sqrt[6]{x}+C