What is the integral of x^5sqrt(4-x^3) ?

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

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integral of x^5sqrt(4-x^3)

\int x^{5} 4 - x^{3} d x

- \frac{1}{3} (- \frac{2}{5} (4 - x^{3})^{\frac{5}{2}} + \frac{8}{3} (4 - x^{3})^{\frac{3}{2}}) + C

Solution steps

\int x^{5} 4 - x^{3} d x

Apply u-substitution

: \int - \frac{( - u + 4 ) u}{3} d u

= \int - \frac{( - u + 4 ) u}{3} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int (- u + 4) u d u

Expand

(- u + 4) u : - u^{\frac{3}{2}} + 4 u

= - \frac{1}{3} \cdot \int - u^{\frac{3}{2}} + 4 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{3} (- \int u^{\frac{3}{2}} d u + \int 4 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= - \frac{1}{3} (- \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = 4 - x^{3}

= - \frac{1}{3} (- \frac{2}{5} (4 - x^{3})^{\frac{5}{2}} + \frac{8}{3} (4 - x^{3})^{\frac{3}{2}})

Add a constant to the solution

= - \frac{1}{3} (- \frac{2}{5} (4 - x^{3})^{\frac{5}{2}} + \frac{8}{3} (4 - x^{3})^{\frac{3}{2}}) + C

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