What is the integral of x^3(7-x^2)^{1/2} ?

The integral of x^3(7-x^2)^{1/2} is -1/15 (3x^2+14)(-x^2+7)^{3/2}+C

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integral of x^3(7-x^2)^{1/2}

\int x^{3} (7 - x^{2})^{\frac{1}{2}} d x

- \frac{1}{15} (3 x^{2} + 14) (- x^{2} + 7)^{\frac{3}{2}} + C

Solution steps

\int x^{3} (7 - x^{2})^{\frac{1}{2}} d x

Apply u-substitution

: \int - \frac{( - u + 7 ) u ^{\frac{1}{2}}}{2} d u

= \int - \frac{( - u + 7 ) u ^{\frac{1}{2}}}{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 + 7) u^{\frac{1}{2}} d u

Expand

(- u + 7) u^{\frac{1}{2}} : - u^{\frac{3}{2}} + 7 u^{\frac{1}{2}}

= - \frac{1}{2} \cdot \int - u^{\frac{3}{2}} + 7 u^{\frac{1}{2}} 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{3}{2}} d u + \int 7 u^{\frac{1}{2}} d u)

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

\int 7 u^{\frac{1}{2}} d u = \frac{14}{3} u^{\frac{3}{2}}

= - \frac{1}{2} (- \frac{2}{5} u^{\frac{5}{2}} + \frac{14}{3} u^{\frac{3}{2}})

Substitute back

u = 7 - x^{2}

= - \frac{1}{2} (- \frac{2}{5} (7 - x^{2})^{\frac{5}{2}} + \frac{14}{3} (7 - x^{2})^{\frac{3}{2}})

Simplify

- \frac{1}{2} (- \frac{2}{5} (7 - x^{2})^{\frac{5}{2}} + \frac{14}{3} (7 - x^{2})^{\frac{3}{2}}) : - \frac{1}{15} (3 x^{2} + 14) (- x^{2} + 7)^{\frac{3}{2}}

= - \frac{1}{15} (3 x^{2} + 14) (- x^{2} + 7)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{15} (3 x^{2} + 14) (- x^{2} + 7)^{\frac{3}{2}} + C

y^{^{'}} = e^{9 x} - 4 x

\int e^{- 2 x^{2}} \cdot (- 4) x^{3} d x

n = 1 \sum \infty 8 (0.2)^{n}

n = 1 \sum \infty (1 + \frac{3}{n})^{n}

x \to - 1 lim (\frac{x ^{3} - 1}{x - 1})

What is the integral of x^3(7-x^2)^{1/2} ?
The integral of x^3(7-x^2)^{1/2} is -1/15 (3x^2+14)(-x^2+7)^{3/2}+C