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

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

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

\int (x - 4) 3 x^{2} - 8 x d x

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

Solution steps

\int (x - 4) 3 x^{2} - 8 x d x

Apply u-substitution

: \int u 3 u^{2} - 16 d u

= \int u 3 u^{2} - 16 d u

Apply u-substitution

: \int \frac{3 v}{2} d v

= \int \frac{3 v}{2} d v

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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