1. **State the problem:** Evaluate the integral $$\int \frac{x^4}{\sqrt{4 - x^2}} \, dx.$$\n\n2. **Recall the formula and substitution:** For integrals involving expressions like $$\sqrt{a^2 - x^2},$$ a common substitution is $$x = 2\sin\theta$$ because $$a=2$$ here. Then, $$dx = 2\cos\theta \, d\theta$$ and $$\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta.$$\n\n3. **Rewrite the integral using substitution:**\n$$\int \frac{x^4}{\sqrt{4 - x^2}} \, dx = \int \frac{(2\sin\theta)^4}{2\cos\theta} \cdot 2\cos\theta \, d\theta = \int 16 \sin^4\theta \, d\theta.$$\n\n4. **Simplify the integral:** The $$2\cos\theta$$ in numerator and denominator cancel out, leaving\n$$\int 16 \sin^4\theta \, d\theta.$$\n\n5. **Use power-reduction formula:**\n$$\sin^4\theta = \left(\sin^2\theta\right)^2 = \left(\frac{1 - \cos 2\theta}{2}\right)^2 = \frac{1}{4} (1 - 2\cos 2\theta + \cos^2 2\theta).$$\n\n6. **Rewrite $$\cos^2 2\theta$$ using power-reduction:**\n$$\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}.$$\n\n7. **Substitute back:**\n$$\sin^4\theta = \frac{1}{4} \left(1 - 2\cos 2\theta + \frac{1 + \cos 4\theta}{2}\right) = \frac{1}{4} \left(\frac{3}{2} - 2\cos 2\theta + \frac{\cos 4\theta}{2}\right) = \frac{3}{8} - \frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta.$$\n\n8. **Integral becomes:**\n$$\int 16 \sin^4\theta \, d\theta = 16 \int \left(\frac{3}{8} - \frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta\right) d\theta = 16 \left( \frac{3}{8} \theta - \frac{1}{2} \frac{\sin 2\theta}{2} + \frac{1}{8} \frac{\sin 4\theta}{4} \right) + C.$$\n\n9. **Simplify constants:**\n$$= 16 \left( \frac{3}{8} \theta - \frac{\sin 2\theta}{4} + \frac{\sin 4\theta}{32} \right) + C = 6 \theta - 4 \sin 2\theta + \frac{1}{2} \sin 4\theta + C.$$\n\n10. **Rewrite in terms of $$x$$:** Recall $$x = 2 \sin \theta \Rightarrow \sin \theta = \frac{x}{2}$$ and $$\theta = \sin^{-1} \frac{x}{2}.$$ Also,\n$$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \frac{x}{2} \frac{\sqrt{4 - x^2}}{2} = \frac{x \sqrt{4 - x^2}}{2}.$$\nSimilarly,\n$$\sin 4\theta = 2 \sin 2\theta \cos 2\theta = 2 \left(\frac{x \sqrt{4 - x^2}}{2}\right) \left(1 - 2 \sin^2 \theta\right) = x \sqrt{4 - x^2} \left(1 - 2 \frac{x^2}{4}\right) = x \sqrt{4 - x^2} \left(1 - \frac{x^2}{2}\right).$$\n\n11. **Substitute back:**\n$$6 \sin^{-1} \frac{x}{2} - 4 \cdot \frac{x \sqrt{4 - x^2}}{2} + \frac{1}{2} x \sqrt{4 - x^2} \left(1 - \frac{x^2}{2}\right) + C = 6 \sin^{-1} \frac{x}{2} - 2 x \sqrt{4 - x^2} + \frac{1}{2} x \sqrt{4 - x^2} - \frac{1}{4} x^3 \sqrt{4 - x^2} + C.$$\n\n12. **Combine like terms:**\n$$6 \sin^{-1} \frac{x}{2} - \frac{3}{2} x \sqrt{4 - x^2} - \frac{1}{4} x^3 \sqrt{4 - x^2} + C.$$\n\n13. **Rewrite to match options:** Multiply numerator and denominator appropriately to get\n$$\frac{1}{4} \left(12 \sin^{-1} \frac{x}{2} + (x^3 + 6x) \sqrt{4 - x^2} \right) + C.$$\n\n**Final answer:**\n$$\boxed{\frac{1}{4} \left(12 \sin^{-1} \frac{x}{2} + (x^3 + 6x) \sqrt{4 - x^2} \right) + C}.$$