What is the integral of 6\sqrt[3]{x} ?

The integral of 6\sqrt[3]{x} is 9/2 x^{4/3}+C

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integral of 6\sqrt[3]{x}

\frac{9}{2} x^{\frac{4}{3}} + C

Solution steps

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} x^{\frac{4}{3}}

= 6 \cdot \frac{3}{4} x^{\frac{4}{3}}

Simplify

6 \cdot \frac{3}{4} x^{\frac{4}{3}} : \frac{9}{2} x^{\frac{4}{3}}

= \frac{9}{2} x^{\frac{4}{3}}

Add a constant to the solution

= \frac{9}{2} x^{\frac{4}{3}} + C

- x^{2} + y = 1, [2, - 2]

\int (r^{2} + r + 1) e^{r} d r

x \to 1 + lim (\frac{x ^{2} - 8 x + 15}{x - 1})

2 x y d x + (x^{2} + 1) d y = 3 x d x

\frac{\partial}{\partial r} (e^{r} cos (θ))

What is the integral of 6\sqrt[3]{x} ?
The integral of 6\sqrt[3]{x} is 9/2 x^{4/3}+C