(i) Let \(u = \sqrt{6-x}\), then \(u^2 = 6-x\) and differentiating gives \(2u \, du = -dx\), or \(dx = -2u \, du\).
Substitute \(x = 6-u^2\) into the integral:
\(I = \int \frac{5}{6-u^2 + u} (-2u) \, du\).
Simplify the integrand:
\(I = \int \frac{-10u}{6-u^2+u} \, du = \int \frac{-10u}{(3-u)(2+u)} \, du\).
Change the limits: when \(x = 2\), \(u = \sqrt{4} = 2\); when \(x = 5\), \(u = \sqrt{1} = 1\).
Thus, \(I = \int_{1}^{2} \frac{10u}{(3-u)(2+u)} \, du\).
(ii) Use partial fraction decomposition:
\(\frac{10u}{(3-u)(2+u)} = \frac{A}{3-u} + \frac{B}{2+u}\).
Multiply through by \((3-u)(2+u)\) and equate coefficients to find \(A = 6\) and \(B = -4\).
Integrate each term separately:
\(\int \frac{6}{3-u} \, du = -6 \ln|3-u|\) and \(\int \frac{-4}{2+u} \, du = -4 \ln|2+u|\).
Combine and evaluate from 1 to 2:
\(I = [-6 \ln|3-u| - 4 \ln|2+u|]_{1}^{2}\).
Calculate:
\(I = [-6 \ln(1) - 4 \ln(4)] - [-6 \ln(2) - 4 \ln(3)]\).
Simplify to find \(I = 2 \ln\left(\frac{9}{2}\right)\).
