What is the integral of \sqrt[3]{6x+1} ?

The integral of \sqrt[3]{6x+1} is 1/8 (6x+1)^{4/3}+C

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integral of \sqrt[3]{6x+1}

\frac{1}{8} (6 x + 1)^{\frac{4}{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}{4} u^{\frac{4}{3}}

= \frac{1}{6} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 6 x + 1

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

Simplify

\frac{1}{6} \cdot \frac{3}{4} (6 x + 1)^{\frac{4}{3}} : \frac{1}{8} (6 x + 1)^{\frac{4}{3}}

= \frac{1}{8} (6 x + 1)^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{8} (6 x + 1)^{\frac{4}{3}} + C

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What is the integral of \sqrt[3]{6x+1} ?
The integral of \sqrt[3]{6x+1} is 1/8 (6x+1)^{4/3}+C