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

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

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

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

Solution steps

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{1}{3} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 2 - 3 x

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

Simplify

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

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

Add a constant to the solution

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

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