What is the integral of 8/(\sqrt[3]{x^2)} ?

The integral of 8/(\sqrt[3]{x^2)} is 24\sqrt[3]{x}+C

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integral of 8/(\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

= 8 \cdot \int \frac{1}{x ^{\frac{2}{3}}} d x

Apply the Power Rule

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

= 8 \cdot 3 x^{\frac{1}{3}}

Simplify

Add a constant to the solution

y^{^{''}} - 2 y^{^{'}} = 6 e^{2 x}

\int \frac{9 x ^{2} + 6 x + 9}{x ^{3} - 1} d x

e^{2 t} sin (2 t j)

y = c cos (t) + t^{2} sin (t)

\frac{\partial}{\partial y} (x + 2 y + 1)

What is the integral of 8/(\sqrt[3]{x^2)} ?
The integral of 8/(\sqrt[3]{x^2)} is 24\sqrt[3]{x}+C