What is the integral of 6/(\sqrt[4]{4x-1)} ?

The integral of 6/(\sqrt[4]{4x-1)} is 2(4x-1)^{3/4}+C

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integral of 6/(\sqrt[4]{4x-1)}

\int \frac{6}{4 4 x - 1} d x

2 (4 x - 1)^{\frac{3}{4}} + C

Solution steps

\int \frac{6}{4 4 x - 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{4 4 x - 1} d x

Apply u-substitution

= 6 \cdot \int \frac{1}{4 4 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{4} \cdot \int \frac{1}{4 u} d u

Apply the Power Rule

= 6 \cdot \frac{1}{4} \cdot \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = 4 x - 1

= 6 \cdot \frac{1}{4} \cdot \frac{4}{3} (4 x - 1)^{\frac{3}{4}}

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 6/(\sqrt[4]{4x-1)} ?
The integral of 6/(\sqrt[4]{4x-1)} is 2(4x-1)^{3/4}+C