What is the integral of (16x^3)/(sqrt(4-x^2)) ?

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

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integral of (16x^3)/(sqrt(4-x^2))

\int \frac{16 x ^{3}}{4 - x ^{2}} d x

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

Solution steps

\int \frac{16 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

= 16 \cdot \int \frac{x ^{3}}{4 - x ^{2}} d x

Apply u-substitution

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

= 16 \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

= 16 (\int u^{2} d u - \int 4 d u)

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

\int 4 d u = 4 u

= 16 (\frac{u ^{3}}{3} - 4 u)

Substitute back

u = 4 - x^{2}

= 16 (\frac{( 4 - x ^{2} ) ^{3}}{3} - 4 4 - x^{2})

Simplify

16 (\frac{( 4 - x ^{2} ) ^{3}}{3} - 4 4 - x^{2}) : 16 (\frac{1}{3} (4 - x^{2})^{\frac{3}{2}} - 4 4 - x^{2})

= 16 (\frac{1}{3} (4 - x^{2})^{\frac{3}{2}} - 4 4 - x^{2})

Add a constant to the solution

= 16 (\frac{1}{3} (4 - x^{2})^{\frac{3}{2}} - 4 4 - x^{2}) + C

\int - x + 4 d x

x 2 x + 1

(x + y)^{2} d x + (2 x y + x^{2} - 4) d y = 0, y (1) = 1

\int x sin (3) x^{2} d x

y^{^{'}} - 4 y = 0

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