What is the integral of (6y)^{1/3} ?

The integral of (6y)^{1/3} is (3*3^{1/3})/(2*2^{2/3)}y^{4/3}+C

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integral of (6y)^{1/3}

\int (6 y)^{\frac{1}{3}} d y

\frac{3 \cdot 3 ^{\frac{1}{3}}}{2 \cdot 2 ^{\frac{2}{3}}} y^{\frac{4}{3}} + C

Solution steps

\int (6 y)^{\frac{1}{3}} d y

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= \int 6^{\frac{1}{3}} y^{\frac{1}{3}} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6^{\frac{1}{3}} \cdot \int y^{\frac{1}{3}} d y

Apply the Power Rule

: \frac{3}{4} y^{\frac{4}{3}}

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

Simplify

6^{\frac{1}{3}} \cdot \frac{3}{4} y^{\frac{4}{3}} : \frac{3 \cdot 3 ^{\frac{1}{3}}}{2 \cdot 2 ^{\frac{2}{3}}} y^{\frac{4}{3}}

= \frac{3 \cdot 3 ^{\frac{1}{3}}}{2 \cdot 2 ^{\frac{2}{3}}} y^{\frac{4}{3}}

Add a constant to the solution

= \frac{3 \cdot 3 ^{\frac{1}{3}}}{2 \cdot 2 ^{\frac{2}{3}}} y^{\frac{4}{3}} + C

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What is the integral of (6y)^{1/3} ?
The integral of (6y)^{1/3} is (3*3^{1/3})/(2*2^{2/3)}y^{4/3}+C