What is the integral of 6/(\sqrt[4]{7-4x)} ?

The integral of 6/(\sqrt[4]{7-4x)} is -2(-4x+7)^{3/4}+C

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integral of 6/(\sqrt[4]{7-4x)}

- 2 (- 4 x + 7)^{\frac{3}{4}} + C

Solution steps

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{4}{3} u^{\frac{3}{4}}

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

Substitute back

u = 7 - 4 x

= 6 (- \frac{1}{4} \cdot \frac{4}{3} (7 - 4 x)^{\frac{3}{4}})

Simplify

6 (- \frac{1}{4} \cdot \frac{4}{3} (7 - 4 x)^{\frac{3}{4}}) : - 2 (- 4 x + 7)^{\frac{3}{4}}

= - 2 (- 4 x + 7)^{\frac{3}{4}}

Add a constant to the solution

= - 2 (- 4 x + 7)^{\frac{3}{4}} + C

10 cos (8 t) + 14 sin (8 t)

x \to 1 lim (2 x^{\frac{3}{2}} - 3 x^{2})

\frac{d}{d x} (7 x^{2} + 8 x)

f (x) = 3 x^{2} + 2 x, at p (- 1, 5)

\frac{d}{d x} (cos (4 x) \cdot 4)

What is the integral of 6/(\sqrt[4]{7-4x)} ?
The integral of 6/(\sqrt[4]{7-4x)} is -2(-4x+7)^{3/4}+C