1. **State the problem:** Evaluate the integral $$\int \frac{dx}{\sqrt{8x - x^2}}$$. 2. **Rewrite the expression inside the square root:** $$8x - x^2 = -(x^2 - 8x) = -(x^2 - 8x + 16 - 16) = -(x - 4)^2 + 16$$. 3. **Substitute:** Let $$u = x - 4$$, so $$x = u + 4$$ and $$dx = du$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int \frac{du}{\sqrt{16 - u^2}}$$. 5. **Recognize the integral form:** This is a standard integral of the form $$\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C$$ where $$a = 4$$. 6. **Apply the formula:** $$\int \frac{du}{\sqrt{16 - u^2}} = \arcsin\left(\frac{u}{4}\right) + C$$. 7. **Back-substitute $$u = x - 4$$:** $$\arcsin\left(\frac{x - 4}{4}\right) + C$$. **Final answer:** $$\int \frac{dx}{\sqrt{8x - x^2}} = \arcsin\left(\frac{x - 4}{4}\right) + C$$.