What is the integral of 1/(\sqrt[3]{4x^2-12x+9)} ?

The integral of 1/(\sqrt[3]{4x^2-12x+9)} is 3/(\sqrt[3]{4)}\sqrt[3]{1/2 (2x-3)}+C

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integral of 1/(\sqrt[3]{4x^2-12x+9)}

Solution steps

Complete the square

4 x^{2} - 12 x + 9 : 4 (x - \frac{3}{2})^{2}

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

: 3 u^{\frac{1}{3}}

Substitute back

u = x - \frac{3}{2}

Simplify

Add a constant to the solution

\int - t d t

\int sin^{2} (x) e^{x} d x

x^{2} sin (x)

\int x (x + 4)^{8} d x

\int e^{s i n (4 x)} cos (4 x) d x

What is the integral of 1/(\sqrt[3]{4x^2-12x+9)} ?
The integral of 1/(\sqrt[3]{4x^2-12x+9)} is 3/(\sqrt[3]{4)}\sqrt[3]{1/2 (2x-3)}+C