The problem asks us to evaluate the indefinite integral $I = \int \frac{x^3}{\sqrt{4-x^2}} dx$.

1. Problem Description

The problem asks us to evaluate the indefinite integral

$I = \int \frac{x^3}{\sqrt{4-x^2}} dx$.

2. Solution Steps

To solve the integral $I = \int \frac{x^3}{\sqrt{4-x^2}} dx$, we use u-substitution.

Let $u = 4 - x^2$. Then $du = -2x dx$, so $x dx = -\frac{1}{2} du$.

Also, $x^2 = 4 - u$. Then we can rewrite the integral as

$I = \int \frac{x^2}{\sqrt{4-x^2}} x dx = \int \frac{4-u}{\sqrt{u}} (-\frac{1}{2}) du = -\frac{1}{2} \int \frac{4-u}{\sqrt{u}} du$.

$I = -\frac{1}{2} \int (\frac{4}{\sqrt{u}} - \frac{u}{\sqrt{u}}) du = -\frac{1}{2} \int (4u^{-1/2} - u^{1/2}) du$.

$I = -\frac{1}{2} (4 \cdot \frac{u^{1/2}}{1/2} - \frac{u^{3/2}}{3/2}) + C = -\frac{1}{2} (8u^{1/2} - \frac{2}{3} u^{3/2}) + C$.

$I = -4u^{1/2} + \frac{1}{3} u^{3/2} + C$.

Substitute $u = 4 - x^2$:

$I = -4\sqrt{4-x^2} + \frac{1}{3} (4-x^2)^{3/2} + C = -4\sqrt{4-x^2} + \frac{1}{3} (4-x^2)\sqrt{4-x^2} + C$.

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

$I = \sqrt{4-x^2}(\frac{-12+4}{3} - \frac{1}{3}x^2) + C = \sqrt{4-x^2}(\frac{-8}{3} - \frac{1}{3}x^2) + C$.

$I = -\frac{1}{3}(8+x^2)\sqrt{4-x^2} + C$.

3. Final Answer

$I = -\frac{1}{3}(x^2+8)\sqrt{4-x^2} + C$