What is the integral of 2x^2sqrt(1-x^3) ?

The integral of 2x^2sqrt(1-x^3) is -4/9 (1-x^3)^{3/2}+C

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integral of 2x^2sqrt(1-x^3)

\int 2 x^{2} 1 - x^{3} d x

- \frac{4}{9} (1 - x^{3})^{\frac{3}{2}} + C

Solution steps

\int 2 x^{2} 1 - x^{3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int x^{2} 1 - x^{3} d x

Apply u-substitution

: \int - \frac{u}{3} d u

= 2 \cdot \int - \frac{u}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 (- \frac{1}{3} \cdot \int u d u)

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = 1 - x^{3}

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

Simplify

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

= - \frac{4}{9} (1 - x^{3})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{4}{9} (1 - x^{3})^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (3 x y + y^{2})

\int (2 - 3 e^{x}) d x

\frac{\partial}{\partial x} (cos (x^{2} y^{3}))

\int \frac{x + 1}{x ^{3} - 6 x ^{2} + 9 x} d x

x \to 0 lim (- (\frac{6 x}{x ^{4}}))

What is the integral of 2x^2sqrt(1-x^3) ?
The integral of 2x^2sqrt(1-x^3) is -4/9 (1-x^3)^{3/2}+C