1. **State the problem:** We want to evaluate the integral $$\int_5^{10} \frac{3 \, dx}{(x - 1)(3 + 2\sqrt{x - 1})}$$ using the substitution $$x = u^2 + 1$$ and determine the new limits $$p$$ and $$Q$$ in terms of $$u$$. 2. **Substitution and change of limits:** Given $$x = u^2 + 1$$, then $$dx = 2u \, du$$. When $$x=5$$, $$u = \sqrt{5-1} = \sqrt{4} = 2$$. When $$x=10$$, $$u = \sqrt{10-1} = \sqrt{9} = 3$$. Thus, the new limits are $$p=2$$ and $$Q=3$$. 3. **Rewrite the integral with new limits:** Substitute into the integral: $$\int_5^{10} \frac{3 \, dx}{(x - 1)(3 + 2\sqrt{x - 1})} = \int_2^3 \frac{3 \cdot 2u \, du}{(u^2 + 1 - 1)(3 + 2u)} = \int_2^3 \frac{6u \, du}{u^2 (3 + 2u)}$$ 4. **Simplify the integrand:** Cancel one $$u$$ in numerator and denominator: $$\int_2^3 \frac{6u \, du}{u^2 (3 + 2u)} = \int_2^3 \frac{6 \cancel{u} \, du}{\cancel{u} u (3 + 2u)} = \int_2^3 \frac{6 \, du}{u (3 + 2u)}$$ 5. **Partial fraction decomposition:** Express: $$\frac{6}{u(3+2u)} = \frac{A}{u} + \frac{B}{3+2u}$$ Multiply both sides by $$u(3+2u)$$: $$6 = A(3+2u) + Bu$$ Set $$u=0$$: $$6 = 3A \Rightarrow A=2$$ Set $$u = -\frac{3}{2}$$: $$6 = B \left(-\frac{3}{2}\right) \Rightarrow B = -4$$ 6. **Rewrite integral:** $$\int_2^3 \frac{6}{u(3+2u)} du = \int_2^3 \left( \frac{2}{u} - \frac{4}{3+2u} \right) du$$ 7. **Integrate each term:** $$\int_2^3 \frac{2}{u} du = 2 \ln u \Big|_2^3 = 2 (\ln 3 - \ln 2)$$ For the second term, use substitution $$v = 3 + 2u$$, $$dv = 2 du$$, so $$du = \frac{dv}{2}$$: $$\int_2^3 \frac{4}{3+2u} du = 4 \int_2^3 \frac{1}{3+2u} du = 4 \int_{v=7}^{9} \frac{1}{v} \frac{dv}{2} = 2 \int_7^{9} \frac{1}{v} dv = 2 (\ln 9 - \ln 7)$$ 8. **Combine results:** $$\int_2^3 \frac{6}{u(3+2u)} du = 2 (\ln 3 - \ln 2) - 2 (\ln 9 - \ln 7) = 2 \ln 3 - 2 \ln 2 - 2 \ln 9 + 2 \ln 7 = 2 \ln \left( \frac{3 \cdot 7}{2 \cdot 9} \right) = 2 \ln \frac{21}{18} = 2 \ln \frac{7}{6}$$ 9. **Final answer:** $$\int_5^{10} \frac{3 \, dx}{(x - 1)(3 + 2\sqrt{x - 1})} = 2 \ln \frac{7}{6}$$