What is the integral of x\sqrt[3]{5-6x^2} ?

The integral of x\sqrt[3]{5-6x^2} is -1/16 (5-6x^2)^{4/3}+C

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integral of x\sqrt[3]{5-6x^2}

- \frac{1}{16} (5 - 6 x^{2})^{\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}{12} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 5 - 6 x^{2}

= - \frac{1}{12} \cdot \frac{3}{4} (5 - 6 x^{2})^{\frac{4}{3}}

Simplify

- \frac{1}{12} \cdot \frac{3}{4} (5 - 6 x^{2})^{\frac{4}{3}} : - \frac{1}{16} (5 - 6 x^{2})^{\frac{4}{3}}

= - \frac{1}{16} (5 - 6 x^{2})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{1}{16} (5 - 6 x^{2})^{\frac{4}{3}} + C

\frac{\partial}{\partial y} (tan (2 x + y))

0 = 2 x + 1 - 2 y y^{^{'}}

\int 2 [1 - cos^{2} (2 x)] d x

\int cos (x) (2 + sin (x))^{6} d x

\frac{d y}{d t} + t y^{2} = 0

What is the integral of x\sqrt[3]{5-6x^2} ?
The integral of x\sqrt[3]{5-6x^2} is -1/16 (5-6x^2)^{4/3}+C