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

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

Substitute back

Simplify

Add a constant to the solution

\int \frac{1}{( x ^{2} + 2 x + 10 ) ^{2}} d x

\int \frac{4 x ^{3}}{( x ^{2} + 1 ) ^{3}} d x

\int \frac{4}{x ^{- 3}} d x

\int \frac{3 x ^{2} - 7 x + 5}{( x - 1 ) ( x - 2 ) ^{2}} d x

\int \frac{x}{( 25 + x ^{2} ) ^{3}} d x

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