What is the integral of 1/(\sqrt[4]{6-9x)} ?

The integral of 1/(\sqrt[4]{6-9x)} is -4/27 (6-9x)^{3/4}+C

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integral of 1/(\sqrt[4]{6-9x)}

- \frac{4}{27} (6 - 9 x)^{\frac{3}{4}} + 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{1}{9} \cdot \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = 6 - 9 x

= - \frac{1}{9} \cdot \frac{4}{3} (6 - 9 x)^{\frac{3}{4}}

Simplify

- \frac{1}{9} \cdot \frac{4}{3} (6 - 9 x)^{\frac{3}{4}} : - \frac{4}{27} (6 - 9 x)^{\frac{3}{4}}

= - \frac{4}{27} (6 - 9 x)^{\frac{3}{4}}

Add a constant to the solution

= - \frac{4}{27} (6 - 9 x)^{\frac{3}{4}} + C

What is the integral of 1/(\sqrt[4]{6-9x)} ?
The integral of 1/(\sqrt[4]{6-9x)} is -4/27 (6-9x)^{3/4}+C