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

The integral of (3x^3)/(sqrt(4-x^2)) is (4-x^2)^{3/2}-12sqrt(4-x^2)+C

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

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

(4 - x^{2})^{\frac{3}{2}} - 12 4 - x^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

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

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

\int 4 d u = 4 u

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

Substitute back

u = 4 - x^{2}

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

Simplify

3 (\frac{( 4 - x ^{2} ) ^{3}}{3} - 4 4 - x^{2}) : (4 - x^{2})^{\frac{3}{2}} - 12 4 - x^{2}

= (4 - x^{2})^{\frac{3}{2}} - 12 4 - x^{2}

Add a constant to the solution

= (4 - x^{2})^{\frac{3}{2}} - 12 4 - x^{2} + C

\int \frac{1}{x ^{n}} d x

\int t^{3} e^{- 2 t} d t

\frac{d}{d x} (π ln (e))

\int arccot (y) d y

x \to 0 lim (\frac{3 x ^{2} - x}{x})

What is the integral of (3x^3)/(sqrt(4-x^2)) ?
The integral of (3x^3)/(sqrt(4-x^2)) is (4-x^2)^{3/2}-12sqrt(4-x^2)+C