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

The integral of 2x^2\sqrt[3]{3x^3+4} is 1/6 (3x^3+4)^{4/3}+C

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

\frac{1}{6} (3 x^{3} + 4)^{\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

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}}

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

Substitute back

= 2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{3}{4} (3 x^{3} + 4)^{\frac{4}{3}}

Simplify

2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{3}{4} (3 x^{3} + 4)^{\frac{4}{3}} : \frac{1}{6} (3 x^{3} + 4)^{\frac{4}{3}}

= \frac{1}{6} (3 x^{3} + 4)^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{6} (3 x^{3} + 4)^{\frac{4}{3}} + C

- 0.6 t^{3} + 7.2 t^{2} + 61

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

\int - 3 sin^{2} (x) d x

y = x^{- 5 c o s (x)}

x ln (x) - x

What is the integral of 2x^2\sqrt[3]{3x^3+4} ?
The integral of 2x^2\sqrt[3]{3x^3+4} is 1/6 (3x^3+4)^{4/3}+C