1. **State the problem:** We want to evaluate the definite integral $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1 + 9 \cot(x)}{9 - \cot(x)} \, dx.$$\n\n2. **Recall the cotangent function:** $\cot(x) = \frac{\cos(x)}{\sin(x)}$. The integral involves a rational function of $\cot(x)$.\n\n3. **Substitution:** Let $t = \cot(x)$. Then, $dt = -\csc^2(x) \, dx$. Also, $\csc^2(x) = 1 + \cot^2(x) = 1 + t^2$, so $dx = -\frac{dt}{1 + t^2}$.\n\n4. **Change the limits:** When $x = \frac{\pi}{4}$, $t = \cot(\frac{\pi}{4}) = 1$. When $x = \frac{\pi}{2}$, $t = \cot(\frac{\pi}{2}) = 0$.\n\n5. **Rewrite the integral:**\n$$\int_{\pi/4}^{\pi/2} \frac{1 + 9t}{9 - t} \, dx = \int_{t=1}^{0} \frac{1 + 9t}{9 - t} \left(-\frac{1}{1 + t^2}\right) dt = \int_0^1 \frac{1 + 9t}{9 - t} \frac{1}{1 + t^2} dt.$$\n\n6. **Simplify the integrand:**\n$$\frac{1 + 9t}{(9 - t)(1 + t^2)} = \frac{1 + 9t}{9 - t} \cdot \frac{1}{1 + t^2}.$$\n\n7. **Partial fraction decomposition:** Let\n$$\frac{1 + 9t}{(9 - t)(1 + t^2)} = \frac{A}{9 - t} + \frac{Bt + C}{1 + t^2}.$$\nMultiply both sides by $(9 - t)(1 + t^2)$:\n$$1 + 9t = A(1 + t^2) + (Bt + C)(9 - t).$$\n\n8. **Expand and collect terms:**\n$$1 + 9t = A + A t^2 + 9Bt + 9C - Bt^2 - Ct.$$\nGroup by powers of $t$:\n$$1 + 9t = (A + 9C) + (9B - C) t + (A - B) t^2.$$\n\n9. **Equate coefficients:**\n- Constant term: $A + 9C = 1$\n- Coefficient of $t$: $9B - C = 9$\n- Coefficient of $t^2$: $A - B = 0$\n\n10. **Solve the system:** From $A - B = 0$, we get $A = B$. Substitute into the other equations:\n- $A + 9C = 1$\n- $9A - C = 9$ (since $B = A$)\n\nFrom the second: $C = 9A - 9$. Substitute into the first:\n$$A + 9(9A - 9) = 1 \Rightarrow A + 81A - 81 = 1 \Rightarrow 82A = 82 \Rightarrow A = 1.$$\nThen $B = 1$, and $C = 9(1) - 9 = 0$.\n\n11. **Rewrite the integrand:**\n$$\frac{1 + 9t}{(9 - t)(1 + t^2)} = \frac{1}{9 - t} + \frac{t}{1 + t^2}.$$\n\n12. **Integral becomes:**\n$$\int_0^1 \left( \frac{1}{9 - t} + \frac{t}{1 + t^2} \right) dt = \int_0^1 \frac{1}{9 - t} dt + \int_0^1 \frac{t}{1 + t^2} dt.$$\n\n13. **Integrate each term:**\n- For $\int \frac{1}{9 - t} dt$, use substitution $u = 9 - t$, $du = -dt$:\n$$\int \frac{1}{9 - t} dt = -\int \frac{1}{u} du = -\ln|u| + C = -\ln|9 - t| + C.$$\nEvaluate from 0 to 1:\n$$[-\ln(9 - t)]_0^1 = -\ln(8) + \ln(9) = \ln\left(\frac{9}{8}\right).$$\n\n- For $\int \frac{t}{1 + t^2} dt$, use substitution $v = 1 + t^2$, $dv = 2t dt$, so $t dt = \frac{dv}{2}$:\n$$\int \frac{t}{1 + t^2} dt = \frac{1}{2} \int \frac{1}{v} dv = \frac{1}{2} \ln|v| + C = \frac{1}{2} \ln(1 + t^2) + C.$$\nEvaluate from 0 to 1:\n$$\frac{1}{2} [\ln(1 + 1^2) - \ln(1 + 0)] = \frac{1}{2} \ln(2).$$\n\n14. **Sum the results:**\n$$\ln\left(\frac{9}{8}\right) + \frac{1}{2} \ln(2) = \ln\left(\frac{9}{8}\right) + \ln\left(2^{\frac{1}{2}}\right) = \ln\left(\frac{9}{8} \times \sqrt{2}\right).$$\n\n15. **Final answer:**\n$$\boxed{\ln\left( \frac{9 \sqrt{2}}{8} \right)}.$$