1. **State the problem:** We want to evaluate the integral $$\int \sqrt{\frac{y - 1}{y - 2}} \, dy.$$\n\n2. **Rewrite the integrand:** The integrand is $$\sqrt{\frac{y - 1}{y - 2}} = \frac{\sqrt{y - 1}}{\sqrt{y - 2}}.$$\n\n3. **Substitution:** Let $$t = \sqrt{y - 2} \implies t^2 = y - 2 \implies y = t^2 + 2.$$\nThen, $$dy = 2t \, dt.$$\nAlso, $$\sqrt{y - 1} = \sqrt{t^2 + 2 - 1} = \sqrt{t^2 + 1}.$$\n\n4. **Rewrite the integral in terms of $t$: $$\int \frac{\sqrt{y - 1}}{\sqrt{y - 2}} \, dy = \int \frac{\sqrt{t^2 + 1}}{t} \cdot 2t \, dt = \int 2 \sqrt{t^2 + 1} \, dt.$$\n\n5. **Simplify the integral:** $$\int 2 \sqrt{t^2 + 1} \, dt = 2 \int \sqrt{t^2 + 1} \, dt.$$\n\n6. **Use the standard formula:** $$\int \sqrt{t^2 + a^2} \, dt = \frac{t}{2} \sqrt{t^2 + a^2} + \frac{a^2}{2} \ln \left| t + \sqrt{t^2 + a^2} \right| + C,$$ where here $a = 1.$\n\n7. **Apply the formula:**\n$$2 \int \sqrt{t^2 + 1} \, dt = 2 \left( \frac{t}{2} \sqrt{t^2 + 1} + \frac{1}{2} \ln \left| t + \sqrt{t^2 + 1} \right| \right) + C = t \sqrt{t^2 + 1} + \ln \left| t + \sqrt{t^2 + 1} \right| + C.$$\n\n8. **Back-substitute $t = \sqrt{y - 2}$:**\n$$\boxed{\sqrt{y - 2} \sqrt{y - 1} + \ln \left| \sqrt{y - 2} + \sqrt{y - 1} \right| + C}.$$\n\nThis is the final answer.