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

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

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

- \frac{3}{8} (1 - 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}{2} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int cos (x) sin (x) d x

x \to - 5 lim (\frac{∣ x + 5 ∣}{x + 5})

y = \frac{4}{x}, y = \frac{4}{x ^{2}}, x = 5

\int 1 + \frac{1}{x} d x

\frac{d}{d x} (6 e^{5 x} + e^{- x^{2}})

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