1. **State the problem:** Evaluate the integral $$\int \frac{1}{\sqrt{4-(x+2)}} \, dx$$. 2. **Rewrite the integral:** Notice the expression inside the square root is $$4-(x+2)$$, which simplifies to $$4 - x - 2 = 2 - x$$. So the integral becomes: $$\int \frac{1}{\sqrt{2 - x}} \, dx$$. 3. **Use substitution:** Let $$u = 2 - x$$, then $$du = -dx$$ or $$dx = -du$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int \frac{1}{\sqrt{u}} (-du) = - \int u^{-\frac{1}{2}} \, du$$. 5. **Integrate:** Recall the formula $$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$ for $$n \neq -1$$. Here, $$n = -\frac{1}{2}$$, so: $$- \int u^{-\frac{1}{2}} \, du = - \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = - 2 u^{\frac{1}{2}} + C$$. 6. **Substitute back $$u = 2 - x$$:** $$- 2 \sqrt{2 - x} + C$$. **Final answer:** $$\int \frac{1}{\sqrt{4-(x+2)}} \, dx = - 2 \sqrt{2 - x} + C$$.