What is the integral of 4\sqrt[3]{y+1} ?

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

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integral of 4\sqrt[3]{y+1}

3 (y + 1)^{\frac{4}{3}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Apply the Power Rule

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

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

Substitute back

u = y + 1

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

Simplify

4 \cdot \frac{3}{4} (y + 1)^{\frac{4}{3}} : 3 (y + 1)^{\frac{4}{3}}

= 3 (y + 1)^{\frac{4}{3}}

Add a constant to the solution

= 3 (y + 1)^{\frac{4}{3}} + C

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