What is the integral of \sqrt[3]{3x-4} ?

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

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integral of \sqrt[3]{3x-4}

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

Solution steps

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

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

Substitute back

u = 3 x - 4

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

Simplify

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

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

Add a constant to the solution

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

\int 5 sin^{3} (3 x) cos^{3} (3 x) d x

y = \frac{4}{3}, y = \frac{1}{9 - x ^{2}}

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

y^{^{'}} = 1 - x + 4 y

\frac{\partial}{\partial x} (5 e^{x^{6} y})

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