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

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

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

- \frac{1}{2} (3 - 2 x)^{\frac{2}{3}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= \frac{2}{3} (- \frac{1}{2} \cdot \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 3 - 2 x

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

Simplify

\frac{2}{3} (- \frac{1}{2} \cdot \frac{3}{2} (3 - 2 x)^{\frac{2}{3}}) : - \frac{1}{2} (3 - 2 x)^{\frac{2}{3}}

= - \frac{1}{2} (3 - 2 x)^{\frac{2}{3}}

Add a constant to the solution

= - \frac{1}{2} (3 - 2 x)^{\frac{2}{3}} + C

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