What is the integral of 10x\sqrt[3]{x^2-2} ?

The integral of 10x\sqrt[3]{x^2-2} is 15/4 (x^2-2)^{4/3}+C

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integral of 10x\sqrt[3]{x^2-2}

\int 10 x 3 x^{2} - 2 d x

\frac{15}{4} (x^{2} - 2)^{\frac{4}{3}} + C

Solution steps

\int 10 x 3 x^{2} - 2 d x

Take the constant out:

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

= 10 \cdot \int x 3 x^{2} - 2 d x

Apply u-substitution

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

= 10 \cdot \int \frac{3 u - 2}{2} d u

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \int 3 u - 2 d u

Apply u-substitution

: \int 3 v d v

= 10 \cdot \frac{1}{2} \cdot \int 3 v d v

Apply the Power Rule

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

= 10 \cdot \frac{1}{2} \cdot \frac{3}{4} v^{\frac{4}{3}}

Substitute back

= 10 \cdot \frac{1}{2} \cdot \frac{3}{4} (x^{2} - 2)^{\frac{4}{3}}

Simplify

10 \cdot \frac{1}{2} \cdot \frac{3}{4} (x^{2} - 2)^{\frac{4}{3}} : \frac{15}{4} (x^{2} - 2)^{\frac{4}{3}}

= \frac{15}{4} (x^{2} - 2)^{\frac{4}{3}}

Add a constant to the solution

= \frac{15}{4} (x^{2} - 2)^{\frac{4}{3}} + C

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What is the integral of 10x\sqrt[3]{x^2-2} ?
The integral of 10x\sqrt[3]{x^2-2} is 15/4 (x^2-2)^{4/3}+C