What is the integral of 1/(\sqrt[3]{8-x)} ?

The integral of 1/(\sqrt[3]{8-x)} is -3/2 (8-x)^{2/3}+C

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integral of 1/(\sqrt[3]{8-x)}

\int \frac{1}{3 8 - x} d x

- \frac{3}{2} (8 - x)^{\frac{2}{3}} + C

Solution steps

\int \frac{1}{3 8 - x} d x

Apply u-substitution

: \int - \frac{1}{3 u} d u

= \int - \frac{1}{3 u} d u

Take the constant out:

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

= - \int \frac{1}{3 u} d u

Apply the Power Rule

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

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

Substitute back

u = 8 - x

= - \frac{3}{2} (8 - x)^{\frac{2}{3}}

Add a constant to the solution

= - \frac{3}{2} (8 - x)^{\frac{2}{3}} + C

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What is the integral of 1/(\sqrt[3]{8-x)} ?
The integral of 1/(\sqrt[3]{8-x)} is -3/2 (8-x)^{2/3}+C