What is the integral of 1/(\sqrt[3]{x-4)} ?

The integral of 1/(\sqrt[3]{x-4)} is 3/2 (x-4)^{2/3}+C

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integral of 1/(\sqrt[3]{x-4)}

\frac{3}{2} (x - 4)^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Apply the Power Rule

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

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

Substitute back

u = x - 4

= \frac{3}{2} (x - 4)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{2} (x - 4)^{\frac{2}{3}} + C

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What is the integral of 1/(\sqrt[3]{x-4)} ?
The integral of 1/(\sqrt[3]{x-4)} is 3/2 (x-4)^{2/3}+C