1. **Problem:** Evaluate the definite integral $$\int_1^2 \frac{24}{t^4 - 6t^3} \, dt$$ 2. **Rewrite the denominator:** Factor the denominator: $$t^4 - 6t^3 = t^3(t - 6)$$ 3. **Rewrite the integral:** $$\int_1^2 \frac{24}{t^3(t - 6)} \, dt$$ 4. **Partial fraction decomposition:** Assume $$\frac{24}{t^3(t - 6)} = \frac{A}{t - 6} + \frac{B}{t} + \frac{C}{t^2} + \frac{D}{t^3}$$ Multiply both sides by $$t^3(t - 6)$$: $$24 = A t^3 + B t^2 (t - 6) + C t (t - 6) + D (t - 6)$$ 5. **Expand and collect terms:** $$24 = A t^3 + B t^3 - 6 B t^2 + C t^2 - 6 C t + D t - 6 D$$ Group by powers of $$t$$: $$24 = (A + B) t^3 + (-6 B + C) t^2 + (-6 C + D) t - 6 D$$ 6. **Equate coefficients:** For the constant term: $$-6 D = 24 \implies D = -4$$ For $$t$$ term: $$-6 C + D = 0 \implies -6 C - 4 = 0 \implies C = -\frac{2}{3}$$ For $$t^2$$ term: $$-6 B + C = 0 \implies -6 B - \frac{2}{3} = 0 \implies B = -\frac{1}{9}$$ For $$t^3$$ term: $$A + B = 0 \implies A = -B = \frac{1}{9}$$ 7. **Rewrite the integral:** $$\int_1^2 \left( \frac{1/9}{t - 6} - \frac{1/9}{t} - \frac{2/3}{t^2} - \frac{4}{t^3} \right) dt$$ 8. **Integrate term-by-term:** $$\int \frac{1/9}{t - 6} dt = \frac{1}{9} \ln|t - 6|$$ $$\int -\frac{1}{9t} dt = -\frac{1}{9} \ln|t|$$ $$\int -\frac{2}{3 t^2} dt = -\frac{2}{3} \int t^{-2} dt = -\frac{2}{3} \left(-t^{-1}\right) = \frac{2}{3 t}$$ $$\int -\frac{4}{t^3} dt = -4 \int t^{-3} dt = -4 \left(-\frac{1}{2 t^2}\right) = \frac{2}{t^2}$$ 9. **Evaluate definite integral:** $$\left[ \frac{1}{9} \ln|t - 6| - \frac{1}{9} \ln|t| + \frac{2}{3 t} + \frac{2}{t^2} \right]_1^2$$ 10. **Calculate at bounds:** At $$t=2$$: $$\frac{1}{9} \ln|2 - 6| - \frac{1}{9} \ln 2 + \frac{2}{3 \times 2} + \frac{2}{2^2} = \frac{1}{9} \ln 4 - \frac{1}{9} \ln 2 + \frac{1}{3} + \frac{1}{2}$$ Since $$\ln 4 = 2 \ln 2$$, this is: $$\frac{1}{9} (2 \ln 2) - \frac{1}{9} \ln 2 + \frac{1}{3} + \frac{1}{2} = \frac{1}{9} \ln 2 + \frac{1}{3} + \frac{1}{2}$$ At $$t=1$$: $$\frac{1}{9} \ln|1 - 6| - \frac{1}{9} \ln 1 + \frac{2}{3 \times 1} + \frac{2}{1^2} = \frac{1}{9} \ln 5 + 0 + \frac{2}{3} + 2$$ 11. **Subtract:** $$\left( \frac{1}{9} \ln 2 + \frac{1}{3} + \frac{1}{2} \right) - \left( \frac{1}{9} \ln 5 + \frac{2}{3} + 2 \right) = \frac{1}{9} (\ln 2 - \ln 5) + \left( \frac{1}{3} - \frac{2}{3} \right) + \left( \frac{1}{2} - 2 \right)$$ Simplify: $$\frac{1}{9} \ln \frac{2}{5} - \frac{1}{3} - \frac{3}{2} = \frac{1}{9} \ln \frac{2}{5} - \frac{11}{6}$$ **Final answer:** $$\boxed{\frac{1}{9} \ln \frac{2}{5} - \frac{11}{6}}$$