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

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

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

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

Substitute back

u = 3 - x

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

Add a constant to the solution

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

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