1. **State the problem:** We want to use the substitution $x = u^2 + 1$ to show that $$\int_5^{10} \frac{3 \, dx}{(x - 1)(3 + 2\sqrt{x - 1})} = \int_p^q \frac{6 \, du}{u(3 + 2u)}$$ 2. **Substitution and limits:** Given $x = u^2 + 1$, then $$x - 1 = u^2$$ When $x = 5$, $$5 = u^2 + 1 \implies u^2 = 4 \implies u = 2$$ When $x = 10$, $$10 = u^2 + 1 \implies u^2 = 9 \implies u = 3$$ So the new limits are $p = 2$ and $q = 3$. 3. **Find $dx$ in terms of $du$:** $$dx = \frac{d}{du}(u^2 + 1) du = 2u \, du$$ 4. **Rewrite the integral:** Substitute $x - 1 = u^2$ and $dx = 2u \, du$ into the integral: $$\int_5^{10} \frac{3 \, dx}{(x - 1)(3 + 2\sqrt{x - 1})} = \int_2^3 \frac{3 \cdot 2u \, du}{u^2 (3 + 2u)}$$ 5. **Simplify the integrand:** $$= \int_2^3 \frac{6u}{u^2 (3 + 2u)} \, du = \int_2^3 \frac{6u}{u^2 (3 + 2u)} \, du$$ Cancel one $u$ from numerator and denominator: $$= \int_2^3 \frac{6 \cancel{u}}{\cancel{u} u (3 + 2u)} \, du = \int_2^3 \frac{6}{u (3 + 2u)} \, du$$ 6. **Final result:** We have shown that $$\int_5^{10} \frac{3 \, dx}{(x - 1)(3 + 2\sqrt{x - 1})} = \int_2^3 \frac{6 \, du}{u (3 + 2u)}$$ Thus, $p = 2$ and $q = 3$. This completes the substitution and verification.