What is the integral of x^3(4-x^2)^{1/4} ?

The integral of x^3(4-x^2)^{1/4} is -2/45 (5x^2+16)(-x^2+4)^{5/4}+C

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integral of x^3(4-x^2)^{1/4}

\int x^{3} (4 - x^{2})^{\frac{1}{4}} d x

- \frac{2}{45} (5 x^{2} + 16) (- x^{2} + 4)^{\frac{5}{4}} + C

Solution steps

\int x^{3} (4 - x^{2})^{\frac{1}{4}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

(- u + 4) u^{\frac{1}{4}} : - u^{\frac{5}{4}} + 4 u^{\frac{1}{4}}

= - \frac{1}{2} \cdot \int - u^{\frac{5}{4}} + 4 u^{\frac{1}{4}} 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}{2} (- \int u^{\frac{5}{4}} d u + \int 4 u^{\frac{1}{4}} d u)

\int u^{\frac{5}{4}} d u = \frac{4}{9} u^{\frac{9}{4}}

\int 4 u^{\frac{1}{4}} d u = \frac{16}{5} u^{\frac{5}{4}}

= - \frac{1}{2} (- \frac{4}{9} u^{\frac{9}{4}} + \frac{16}{5} u^{\frac{5}{4}})

Substitute back

u = 4 - x^{2}

= - \frac{1}{2} (- \frac{4}{9} (4 - x^{2})^{\frac{9}{4}} + \frac{16}{5} (4 - x^{2})^{\frac{5}{4}})

Simplify

- \frac{1}{2} (- \frac{4}{9} (4 - x^{2})^{\frac{9}{4}} + \frac{16}{5} (4 - x^{2})^{\frac{5}{4}}) : - \frac{2}{45} (5 x^{2} + 16) (- x^{2} + 4)^{\frac{5}{4}}

= - \frac{2}{45} (5 x^{2} + 16) (- x^{2} + 4)^{\frac{5}{4}}

Add a constant to the solution

= - \frac{2}{45} (5 x^{2} + 16) (- x^{2} + 4)^{\frac{5}{4}} + C

\int_{0}^{1} x^{2} (x + 1)^{6} d x

\frac{6}{s ( s + 3 )}

\frac{d}{d x} (\frac{1}{1 + 5 x})

\frac{d}{d x} (a sin^{3} (x))

x \to 0 - lim (\frac{14 x ^{3} + 15 x}{x})

What is the integral of x^3(4-x^2)^{1/4} ?
The integral of x^3(4-x^2)^{1/4} is -2/45 (5x^2+16)(-x^2+4)^{5/4}+C