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

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

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integral of (3x^2)/(\sqrt[3]{x^3-2)}

\frac{3}{2} (x^{3} - 2)^{\frac{2}{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

Apply the Power Rule

: \frac{3}{2} v^{\frac{2}{3}}

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

Substitute back

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

Simplify

3 \cdot \frac{1}{3} \cdot \frac{3}{2} (x^{3} - 2)^{\frac{2}{3}} : \frac{3}{2} (x^{3} - 2)^{\frac{2}{3}}

= \frac{3}{2} (x^{3} - 2)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{2} (x^{3} - 2)^{\frac{2}{3}} + C

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