1. **State the problem:** We are given the integral $$\int_5^{10} \frac{3 \, dx}{(x-1)(3 + 2\sqrt{x-1})}$$ and it is equal to $$\int_p^q \frac{6 \, du}{u(3 + 2u)}$$ where $p$ and $q$ are positive constants to be found. We want to find $p$, $q$, and then evaluate the integral to show it equals $\ln a$ for some rational constant $a$. 2. **Find substitution and constants $p$, $q$:** Let $$u = \sqrt{x-1} \implies u^2 = x-1 \implies x = u^2 + 1$$ Then, $$dx = 2u \, du$$ Change the limits: When $x=5$, $$u = \sqrt{5-1} = \sqrt{4} = 2$$ When $x=10$, $$u = \sqrt{10-1} = \sqrt{9} = 3$$ So, $$p = 2, \quad q = 3$$ 3. **Rewrite the integral in terms of $u$:** Substitute into the integral: $$\int_5^{10} \frac{3 \, dx}{(x-1)(3 + 2\sqrt{x-1})} = \int_2^3 \frac{3 \cdot 2u \, du}{(u^2)(3 + 2u)} = \int_2^3 \frac{6u \, du}{u^2(3 + 2u)}$$ Simplify the $u$ terms: $$= \int_2^3 \frac{6u \, du}{u^2(3 + 2u)} = \int_2^3 \frac{6 \, du}{u(3 + 2u)}$$ This matches the given integral on the right side, confirming $p=2$ and $q=3$. 4. **Evaluate the integral algebraically:** We want to evaluate $$I = \int_2^3 \frac{6}{u(3 + 2u)} \, du$$ Use partial fractions: Assume $$\frac{6}{u(3 + 2u)} = \frac{A}{u} + \frac{B}{3 + 2u}$$ Multiply both sides by $u(3 + 2u)$: $$6 = A(3 + 2u) + Bu$$ $$6 = 3A + 2Au + Bu = 3A + u(2A + B)$$ Equate coefficients: - Constant term: $3A = 6 \implies A = 2$ - Coefficient of $u$: $2A + B = 0 \implies 2(2) + B = 0 \implies B = -4$ So, $$\frac{6}{u(3 + 2u)} = \frac{2}{u} - \frac{4}{3 + 2u}$$ 5. **Integrate term-by-term:** $$I = \int_2^3 \left( \frac{2}{u} - \frac{4}{3 + 2u} \right) du = 2 \int_2^3 \frac{1}{u} du - 4 \int_2^3 \frac{1}{3 + 2u} du$$ First integral: $$2 \int_2^3 \frac{1}{u} du = 2 [\ln|u|]_2^3 = 2(\ln 3 - \ln 2) = 2 \ln \frac{3}{2}$$ Second integral: Let $w = 3 + 2u$, then $dw = 2 du \implies du = \frac{dw}{2}$ Change limits for $w$: When $u=2$, $w=3+4=7$ When $u=3$, $w=3+6=9$ So, $$\int_2^3 \frac{1}{3 + 2u} du = \int_7^9 \frac{1}{w} \cdot \frac{dw}{2} = \frac{1}{2} \int_7^9 \frac{1}{w} dw = \frac{1}{2} [\ln|w|]_7^9 = \frac{1}{2} (\ln 9 - \ln 7) = \frac{1}{2} \ln \frac{9}{7}$$ Multiply by $-4$: $$-4 \times \frac{1}{2} \ln \frac{9}{7} = -2 \ln \frac{9}{7}$$ 6. **Combine results:** $$I = 2 \ln \frac{3}{2} - 2 \ln \frac{9}{7} = 2 \left( \ln \frac{3}{2} - \ln \frac{9}{7} \right) = 2 \ln \left( \frac{3/2}{9/7} \right) = 2 \ln \left( \frac{3}{2} \times \frac{7}{9} \right) = 2 \ln \frac{7}{6}$$ 7. **Simplify final expression:** $$I = 2 \ln \frac{7}{6} = \ln \left( \frac{7}{6} \right)^2 = \ln \frac{49}{36}$$ Thus, $$\int_5^{10} \frac{3 \, dx}{(x-1)(3 + 2\sqrt{x-1})} = \ln \frac{49}{36}$$ So the rational constant $a$ is $$a = \frac{49}{36}$$ --- **Final answers:** - $p = 2$ - $q = 3$ - $a = \frac{49}{36}$ ---