What is the integral of \sqrt[3]{(2+9y)^2} ?

The integral of \sqrt[3]{(2+9y)^2} is 1/15 (2+9y)^{5/3}+C

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integral of \sqrt[3]{(2+9y)^2}

\frac{1}{15} (2 + 9 y)^{\frac{5}{3}} + C

Solution steps

assuming

(2 + 9 y) \geq 0

= \int (2 + 9 y)^{\frac{2}{3}} d y

Apply u-substitution

: \int \frac{u ^{\frac{2}{3}}}{9} d u

= \int \frac{u ^{\frac{2}{3}}}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int u^{\frac{2}{3}} d u

Apply the Power Rule

: \frac{3}{5} u^{\frac{5}{3}}

= \frac{1}{9} \cdot \frac{3}{5} u^{\frac{5}{3}}

Substitute back

u = 2 + 9 y

= \frac{1}{9} \cdot \frac{3}{5} (2 + 9 y)^{\frac{5}{3}}

Simplify

\frac{1}{9} \cdot \frac{3}{5} (2 + 9 y)^{\frac{5}{3}} : \frac{1}{15} (2 + 9 y)^{\frac{5}{3}}

= \frac{1}{15} (2 + 9 y)^{\frac{5}{3}}

Add a constant to the solution

= \frac{1}{15} (2 + 9 y)^{\frac{5}{3}} + C

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What is the integral of \sqrt[3]{(2+9y)^2} ?
The integral of \sqrt[3]{(2+9y)^2} is 1/15 (2+9y)^{5/3}+C