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)}

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

\frac{9}{8} 3 4 x - 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= \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

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

= \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

\frac{3}{2} \cdot \frac{1}{4} \cdot 3 (4 x - 1)^{\frac{1}{3}} : \frac{9}{8} 3 4 x - 1

= \frac{9}{8} 3 4 x - 1

Add a constant to the solution

= \frac{9}{8} 3 4 x - 1 + C

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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