What is the integral of \sqrt[3]{x^2-5x+1}(2x-5) ?

The integral of \sqrt[3]{x^2-5x+1}(2x-5) is 3/(16\sqrt[3]{4)}(4x^2-20x+4)^{4/3}+C

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integral of \sqrt[3]{x^2-5x+1}(2x-5)

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_{0}^{\infty} 3 sin^{2} (x) d x

y = 4 x, y = 4 x^{2}

x \to 0 lim (\frac{x}{- 14 x ^{4} + 5 x})

\int \frac{y ^{2}}{1 + y ^{3}} d y

y = 9 e^{x} cos (x), (0, 9)

What is the integral of \sqrt[3]{x^2-5x+1}(2x-5) ?
The integral of \sqrt[3]{x^2-5x+1}(2x-5) is 3/(16\sqrt[3]{4)}(4x^2-20x+4)^{4/3}+C