1. We are asked to evaluate the integral $$\int \frac{dx}{\sqrt{x^2 - 10x + 41}}.$$\n\n2. First, complete the square inside the square root to simplify the expression.\n\n$$x^2 - 10x + 41 = (x^2 - 10x + 25) + 16 = (x - 5)^2 + 16.$$\n\n3. The integral becomes $$\int \frac{dx}{\sqrt{(x - 5)^2 + 16}}.$$\n\n4. This is a standard integral of the form $$\int \frac{dx}{\sqrt{(x - a)^2 + b^2}} = \ln \left| x - a + \sqrt{(x - a)^2 + b^2} \right| + C,$$ where $a=5$ and $b=4$.\n\n5. Applying the formula, the integral evaluates to\n\n$$\ln \left| x - 5 + \sqrt{(x - 5)^2 + 16} \right| + C = \ln \left| x - 5 + \sqrt{x^2 - 10x + 41} \right| + C.$$\n\n6. Therefore, the final answer is\n\n$$\int \frac{dx}{\sqrt{x^2 - 10x + 41}} = \ln \left| x - 5 + \sqrt{x^2 - 10x + 41} \right| + C.$$