What is the integral of (3x+2)\sqrt[3]{3x^2+4x} ?

The integral of (3x+2)\sqrt[3]{3x^2+4x} is 1/(8\sqrt[3]{3)}(9x^2+12x)^{4/3}+C

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integral of (3x+2)\sqrt[3]{3x^2+4x}

Solution steps

Apply u-substitution

Take the constant out:

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

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

Substitute back

Simplify

Add a constant to the solution

\frac{d}{d x} (ln (x^{y}))

\int u^{- 2} (2 u^{4} + 3 u^{2} - 2 u^{- 3}) d u

\frac{d}{d x} (\frac{3 x - 4}{2 x ^{2} + 5})

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

x \to \frac{1}{2} lim (∣ 2 x + 3 ∣)

What is the integral of (3x+2)\sqrt[3]{3x^2+4x} ?
The integral of (3x+2)\sqrt[3]{3x^2+4x} is 1/(8\sqrt[3]{3)}(9x^2+12x)^{4/3}+C