(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{6x^2 + 8x + 9}{(2-x)(3+2x)^2} = \frac{A}{2-x} + \frac{B}{3+2x} + \frac{C}{(3+2x)^2}\).

Multiply through by the denominator \((2-x)(3+2x)^2\) to clear the fractions:

\(6x^2 + 8x + 9 = A(3+2x)^2 + B(2-x)(3+2x) + C(2-x)\).

Expand and equate coefficients to solve for \(A, B,\) and \(C\):

\(A = 1, \; B = -1, \; C = 3\).

Thus, \(f(x) = \frac{1}{2-x} - \frac{1}{3+2x} + \frac{3}{(3+2x)^2}\).

(ii) Evaluate \(\int_{-1}^{0} f(x) \, dx\):

\(\int_{-1}^{0} \left( \frac{1}{2-x} - \frac{1}{3+2x} + \frac{3}{(3+2x)^2} \right) \, dx\).

Integrate each term separately:

\(-\ln|2-x| - \frac{1}{2} \ln|3+2x| - \frac{3}{2(3+2x)}\).

Evaluate from \(-1\) to \(0\):

\(\left[ -\ln|2-x| - \frac{1}{2} \ln|3+2x| - \frac{3}{2(3+2x)} \right]_{-1}^{0} = 1 + \frac{1}{2} \ln \left( \frac{3}{4} \right)\).