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

The integral of 3/(2\sqrt[3]{(4x-1)^2)} is 9/8 \sqrt[3]{4x-1}+C

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

Solution steps

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply u-substitution

= \frac{3}{2} \cdot \int \frac{1}{4 u ^{\frac{2}{3}}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{3}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

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

Substitute back

u = 4 x - 1

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

Simplify

Add a constant to the solution

What is the integral of 3/(2\sqrt[3]{(4x-1)^2)} ?
The integral of 3/(2\sqrt[3]{(4x-1)^2)} is 9/8 \sqrt[3]{4x-1}+C