What is the integral of (x^3)/(sqrt(4-x^2)) ?

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

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integral of (x^3)/(sqrt(4-x^2))

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} - 4 d u

= \int u^{2} - 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

= \int u^{2} d u - \int 4 d u

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

\int 4 d u = 4 u

= \frac{u ^{3}}{3} - 4 u

Substitute back

u = 4 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

(- 3, 1), (2, - 1)

\frac{d}{d x} (lo g_{10} (x^{2}))

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

\frac{d}{d x} (0.6 x^{2})

\frac{x}{1 - x ^{2}}

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