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

The integral of \sqrt[3]{x^2-3x}(2x-3) is 3/(16\sqrt[3]{4)}(4x^2-12x)^{4/3}+C

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

Solution steps

Apply u-substitution

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

Substitute back

Simplify

Add a constant to the solution

What is the integral of \sqrt[3]{x^2-3x}(2x-3) ?
The integral of \sqrt[3]{x^2-3x}(2x-3) is 3/(16\sqrt[3]{4)}(4x^2-12x)^{4/3}+C