What is the integral of-1/(2\sqrt[3]{x^2)} ?

The integral of-1/(2\sqrt[3]{x^2)} is -3/2 \sqrt[3]{x}+C

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integral of-1/(2\sqrt[3]{x^2)}

Solution steps

Take the constant out:

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

assuming

x \geq 0

= - \frac{1}{2} \cdot \int \frac{1}{x ^{\frac{2}{3}}} d x

Apply the Power Rule

: 3 x^{\frac{1}{3}}

= - \frac{1}{2} \cdot 3 x^{\frac{1}{3}}

Simplify

Add a constant to the solution

y^{^{'}} = y (4 - t y)

\frac{d}{d x} (cos^{2} (x) + sin^{2} (x))

\int x^{2} cos (8 x) d x

f (x) = \frac{56}{x ^{9}}

What is the integral of-1/(2\sqrt[3]{x^2)} ?
The integral of-1/(2\sqrt[3]{x^2)} is -3/2 \sqrt[3]{x}+C