What is the integral of 3y\sqrt[3]{2y^2-8} ?

The integral of 3y\sqrt[3]{2y^2-8} is 9/16 (2y^2-8)^{4/3}+C

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integral of 3y\sqrt[3]{2y^2-8}

\int 3 y 3 2 y^{2} - 8 d y

\frac{9}{16} (2 y^{2} - 8)^{\frac{4}{3}} + C

Solution steps

\int 3 y 3 2 y^{2} - 8 d y

Take the constant out:

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

= 3 \cdot \int y 3 2 y^{2} - 8 d y

Apply u-substitution

: \int \frac{3 2 u - 8}{2} d u

= 3 \cdot \int \frac{3 2 u - 8}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int 3 2 u - 8 d u

Apply u-substitution

: \int \frac{3 v}{2} d v

= 3 \cdot \frac{1}{2} \cdot \int \frac{3 v}{2} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int 3 v d v

Apply the Power Rule

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

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

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} (2 y^{2} - 8)^{\frac{4}{3}}

Simplify

3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} (2 y^{2} - 8)^{\frac{4}{3}} : \frac{9}{16} (2 y^{2} - 8)^{\frac{4}{3}}

= \frac{9}{16} (2 y^{2} - 8)^{\frac{4}{3}}

Add a constant to the solution

= \frac{9}{16} (2 y^{2} - 8)^{\frac{4}{3}} + C

\frac{d}{d x} (e^{2 (x - 1)^{2}})

f (x) = 0.3 x^{1.2}

a - b y^{2} = \frac{d y}{d x}

(3.9) (- 5.3)

\frac{d}{d t} (- t^{2})

What is the integral of 3y\sqrt[3]{2y^2-8} ?
The integral of 3y\sqrt[3]{2y^2-8} is 9/16 (2y^2-8)^{4/3}+C