What is the integral of (24x)/(\sqrt[3]{3x^2-4)} ?

The integral of (24x)/(\sqrt[3]{3x^2-4)} is 6(3x^2-4)^{2/3}+C

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integral of (24x)/(\sqrt[3]{3x^2-4)}

\int \frac{24 x}{3 3 x ^{2} - 4} d x

6 (3 x^{2} - 4)^{\frac{2}{3}} + C

Solution steps

\int \frac{24 x}{3 3 x ^{2} - 4} d x

Take the constant out:

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

= 24 \cdot \int \frac{x}{3 3 x ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{2 3 3 u - 4} d u

= 24 \cdot \int \frac{1}{2 3 3 u - 4} d u

Take the constant out:

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

= 24 \cdot \frac{1}{2} \cdot \int \frac{1}{3 3 u - 4} d u

Apply u-substitution

: \int \frac{1}{3 3 v} d v

= 24 \cdot \frac{1}{2} \cdot \int \frac{1}{3 3 v} d v

Take the constant out:

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

= 24 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{1}{3 v} d v

Apply the Power Rule

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

= 24 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{2} v^{\frac{2}{3}}

Substitute back

= 24 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{2} (3 x^{2} - 4)^{\frac{2}{3}}

Simplify

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

= 6 (3 x^{2} - 4)^{\frac{2}{3}}

Add a constant to the solution

= 6 (3 x^{2} - 4)^{\frac{2}{3}} + C

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What is the integral of (24x)/(\sqrt[3]{3x^2-4)} ?
The integral of (24x)/(\sqrt[3]{3x^2-4)} is 6(3x^2-4)^{2/3}+C