What is the integral of 3x\sqrt[3]{2-x^2} ?

The integral of 3x\sqrt[3]{2-x^2} is -9/8 (2-x^2)^{4/3}+C

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integral of 3x\sqrt[3]{2-x^2}

- \frac{9}{8} (2 - x^{2})^{\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 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}}

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

Substitute back

u = 2 - x^{2}

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

Simplify

3 (- \frac{1}{2} \cdot \frac{3}{4} (2 - x^{2})^{\frac{4}{3}}) : - \frac{9}{8} (2 - x^{2})^{\frac{4}{3}}

= - \frac{9}{8} (2 - x^{2})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{9}{8} (2 - x^{2})^{\frac{4}{3}} + C

\int (x + 2) x + 1 d x

y^{^{''}} = y + 4 t^{2}

x \to 5 lim (- 3 x)

y = 7 e^{- 6 x}, (0, 7)

\frac{d y}{d x} = 9 x - \frac{y}{x}

What is the integral of 3x\sqrt[3]{2-x^2} ?
The integral of 3x\sqrt[3]{2-x^2} is -9/8 (2-x^2)^{4/3}+C