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To solve the integral \(\int \frac{x \, dx}{\sqrt{4 - x^2}}\), let's proceed step-by-step:

1. **Substitution**: Let \(u = 4 - x^2\). Then, \(du = -2x \, dx\), or \(\frac{du}{-2} = x \, dx\).
   
2. **Rewrite the integral**: Substitute into the integral:
   \[
   \int \frac{x \, dx}{\sqrt{4 - x^2}} = \int \frac{\frac{du}{-2}}{\sqrt{u}} = -\frac{1}{2} \int \frac{du}{\sqrt{u}}.
   \]

3. **Simplify and integrate**:
   \[
   -\frac{1}{2} \int u^{-\frac{1}{2}} \, du = -\frac{1}{2} \cdot 2\sqrt{u} + C = -\sqrt{u} + C.
   \]

4. **Substitute back for \(u\)**:
   \[
   -\sqrt{4 - x^2} + C.
   \]

The correct answer is **(a) \(-\sqrt{4 - x^2} + C\)**.

Would you like more details on any step or have any questions?

### Related Questions:
1. What is the method for solving integrals with square roots in the denominator?
2. How can trigonometric substitutions be used for integrals like this one?
3. What is the general formula for integrating functions of the form \(\int \frac{x}{\sqrt{a^2 - x^2}} \, dx\)?
4. How do we handle definite integrals when using substitution methods?
5. What are common mistakes to avoid when performing substitution in integration?

**Tip**: When integrating, always check if a substitution can simplify the integral into a more familiar form!

¿Cuál es el resultado de integrar \( \int \frac{x \, dx}{\sqrt{4 - x^2}} \)?

Find the solution to the integral \( \int \frac{x \, dx}{\sqrt{4 - x^2}} \) using the substitution method. This step-by-step guide explains key concepts such as integration, substitution, and dealing with square roots in the denominator. Perfect for high school students preparing for calculus.

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