What is the integral of 4/(\sqrt[3]{x^2)} ?

The integral of 4/(\sqrt[3]{x^2)} is 12\sqrt[3]{x}+C

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integral of 4/(\sqrt[3]{x^2)}

\int \frac{4}{3 x ^{2}} d x

12 3 x + C

Solution steps

\int \frac{4}{3 x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{3 x ^{2}} d x

3 x^{2} = x^{\frac{2}{3}},

assuming

x \geq 0

= 4 \cdot \int \frac{1}{x ^{\frac{2}{3}}} d x

Apply the Power Rule

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

= 4 \cdot 3 x^{\frac{1}{3}}

Simplify

4 \cdot 3 x^{\frac{1}{3}} : 12 3 x

= 12 3 x

Add a constant to the solution

= 12 3 x + C

y^{^{''}} - 4 y^{^{'}} - 5 y = - 2, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (e^{3 x} cos (x))

\frac{d}{d x} (\frac{( x ^{2} + 1 )}{x ^{2} - 1})

\int (π - x) cos (π x) d x

x \to 1 lim (\frac{x - 1}{x ^{3} - 1})

What is the integral of 4/(\sqrt[3]{x^2)} ?
The integral of 4/(\sqrt[3]{x^2)} is 12\sqrt[3]{x}+C