What is the integral of 5x*\sqrt[3]{1-x^2} ?

The integral of 5x*\sqrt[3]{1-x^2} is -15/8 (1-x^2)^{4/3}+C

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integral of 5x*\sqrt[3]{1-x^2}

- \frac{15}{8} (1 - 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}}

= 5 (- \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 1 - x^{2}

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

Simplify

5 (- \frac{1}{2} \cdot \frac{3}{4} (1 - x^{2})^{\frac{4}{3}}) : - \frac{15}{8} (1 - x^{2})^{\frac{4}{3}}

= - \frac{15}{8} (1 - x^{2})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{15}{8} (1 - x^{2})^{\frac{4}{3}} + C

\frac{\partial}{\partial x} (\frac{1}{1 - b x})

\int \frac{arctan ( 5 x )}{1 + 25 x ^{2}} d x

\frac{d}{d x} ((2 x - 7)^{4} (x^{2} + x + 1)^{5})

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

x \to \infty lim (- \frac{sin ^{2} ( x ) + 2}{2 x - 5})

What is the integral of 5x*\sqrt[3]{1-x^2} ?
The integral of 5x*\sqrt[3]{1-x^2} is -15/8 (1-x^2)^{4/3}+C