What is the integral of 5(x-4)\sqrt[3]{x^2-8x} ?

The integral of 5(x-4)\sqrt[3]{x^2-8x} is 15/8 (x^2-8x)^{4/3}+C

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integral of 5(x-4)\sqrt[3]{x^2-8x}

\frac{15}{8} (x^{2} - 8 x)^{\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 the Power Rule

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

= 5 \cdot \frac{1}{2} \cdot \frac{3}{4} v^{\frac{4}{3}}

Substitute back

= 5 \cdot \frac{1}{2} \cdot \frac{3}{4} ((x - 4)^{2} - 16)^{\frac{4}{3}}

Simplify

5 \cdot \frac{1}{2} \cdot \frac{3}{4} ((x - 4)^{2} - 16)^{\frac{4}{3}} : \frac{15}{8} (x^{2} - 8 x)^{\frac{4}{3}}

= \frac{15}{8} (x^{2} - 8 x)^{\frac{4}{3}}

Add a constant to the solution

= \frac{15}{8} (x^{2} - 8 x)^{\frac{4}{3}} + C

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What is the integral of 5(x-4)\sqrt[3]{x^2-8x} ?
The integral of 5(x-4)\sqrt[3]{x^2-8x} is 15/8 (x^2-8x)^{4/3}+C