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

The integral of 1/(\sqrt[3]{9x+1)} is 1/6 (9x+1)^{2/3}+C

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

\frac{1}{6} (9 x + 1)^{\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{1}{9} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 9 x + 1

= \frac{1}{9} \cdot \frac{3}{2} (9 x + 1)^{\frac{2}{3}}

Simplify

\frac{1}{9} \cdot \frac{3}{2} (9 x + 1)^{\frac{2}{3}} : \frac{1}{6} (9 x + 1)^{\frac{2}{3}}

= \frac{1}{6} (9 x + 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{6} (9 x + 1)^{\frac{2}{3}} + C

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What is the integral of 1/(\sqrt[3]{9x+1)} ?
The integral of 1/(\sqrt[3]{9x+1)} is 1/6 (9x+1)^{2/3}+C