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

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

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

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

Solution steps

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}}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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