What is the integral of 1/(\sqrt[3]{x-9)} ?

The integral of 1/(\sqrt[3]{x-9)} is 3/2 (x-9)^{2/3}+C

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integral of 1/(\sqrt[3]{x-9)}

\int \frac{1}{3 x - 9} d x

\frac{3}{2} (x - 9)^{\frac{2}{3}} + C

Solution steps

\int \frac{1}{3 x - 9} d x

Apply u-substitution

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

= \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 = x - 9

= \frac{3}{2} (x - 9)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{2} (x - 9)^{\frac{2}{3}} + C

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What is the integral of 1/(\sqrt[3]{x-9)} ?
The integral of 1/(\sqrt[3]{x-9)} is 3/2 (x-9)^{2/3}+C