(i) To express \(f(x)\) in the form \(A + \frac{B}{x} + \frac{Cx + D}{x^2 + 2}\), perform partial fraction decomposition:

1. Set \(f(x) = \frac{3x^3 + 6x - 8}{x(x^2 + 2)} = A + \frac{B}{x} + \frac{Cx + D}{x^2 + 2}\).

2. Multiply through by \(x(x^2 + 2)\) to clear the denominators:

\(3x^3 + 6x - 8 = A(x^2 + 2) + B(x^2 + 2) + (Cx + D)x\).

3. Expand and collect like terms:

\(3x^3 + 6x - 8 = Ax^2 + 2A + Bx^2 + 2B + Cx^2 + Dx\).

4. Equate coefficients:

\(A + B + C = 3\)

\(D = 6\)

\(2A + 2B = -8\)

5. Solve the system of equations to find \(A = 3, B = -4, C = 4, D = 0\).

(ii) To show \(\int_1^2 f(x) \, dx = 3 - \ln 4\):

1. Integrate \(f(x) = 3 - \frac{4}{x} + \frac{4x}{x^2 + 2}\).

2. Integrate each term separately:

\(\int 3 \, dx = 3x\)

\(\int -\frac{4}{x} \, dx = -4 \ln |x|\)

\(\int \frac{4x}{x^2 + 2} \, dx = 2 \ln(x^2 + 2)\)

3. Evaluate from 1 to 2:

\(\left[ 3x - 4 \ln |x| + 2 \ln(x^2 + 2) \right]_1^2\)

4. Substitute the limits:

\((6 - 4 \ln 2 + 2 \ln 6) - (3 - 4 \ln 1 + 2 \ln 3)\)

5. Simplify to obtain \(3 - \ln 4\).