What is the integral of ((x^3))/(sqrt(9-x^2)) ?

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

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integral of ((x^3))/(sqrt(9-x^2))

\int \frac{( x ^{3} )}{9 - x ^{2}} d x

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

Solution steps

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

Apply u-substitution

: \int u^{2} - 9 d u

= \int u^{2} - 9 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

= \int u^{2} d u - \int 9 d u

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

\int 9 d u = 9 u

= \frac{u ^{3}}{3} - 9 u

Substitute back

u = 9 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d ^{2}}{d x ^{2}} (x^{e^{x}})

h (x) = \frac{( 3 x - 5 )}{1 - x ^{2}}

\frac{d}{d x} (3 x^{2})

\frac{d}{d x} (2 x^{3} + \frac{5 x + 1}{3 x - 5} - \frac{2}{x ^{3}})

y^{^{'}} + y = 8 x e^{x}

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