What is the integral of \sqrt[3]{x+9} ?

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

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integral of \sqrt[3]{x+9}

\frac{3}{4} (x + 9)^{\frac{4}{3}} + C

Solution steps

Apply u-substitution

Apply the Power Rule

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

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

Substitute back

u = x + 9

= \frac{3}{4} (x + 9)^{\frac{4}{3}}

Add a constant to the solution

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

\frac{\partial}{\partial x} (cos (3 x + 4 y))

x \to 1 lim (3 - 2 x)

\int sin (x) \cdot sin (2 x) d x

y = x, y = - x + 3, y = 0

\int (sin^{2} (2 x)) d x

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