(i) To express \(f(x)\) in the form \(\frac{A}{2-x} + \frac{Bx+C}{4+x^2}\), we equate:

\(\frac{12 + 8x - x^2}{(2-x)(4+x^2)} = \frac{A}{2-x} + \frac{Bx+C}{4+x^2}\)

Multiply through by \((2-x)(4+x^2)\) to clear the denominators:

\(12 + 8x - x^2 = A(4+x^2) + (Bx+C)(2-x)\)

Expand and collect like terms:

\(12 + 8x - x^2 = 4A + Ax^2 + 2Bx + Cx - Bx^2 - Cx\)

\(= (A-B)x^2 + (2B+C)x + 4A\)

Equating coefficients, we get:

\(A - B = -1\)

\(2B + C = 8\)

\(4A = 12\)

Solving these equations gives \(A = 3\), \(B = 4\), \(C = 0\).

(ii) To show \(\int_0^1 f(x) \, dx = \ln\left(\frac{25}{2}\right)\), integrate each term separately:

\(\int \frac{3}{2-x} \, dx = -3 \ln|2-x|\)

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

\(\int \frac{0}{4+x^2} \, dx = 0\)

Evaluate from 0 to 1:

\(-3 \ln|2-1| + 2 \ln|4+1^2| - (-3 \ln|2-0| + 2 \ln|4+0^2|)\)

\(= -3 \ln(1) + 2 \ln(5) + 3 \ln(2) - 2 \ln(4)\)

\(= 2 \ln(5) + 3 \ln(2) - 2 \ln(4)\)

\(= \ln\left(\frac{5^2 \cdot 2^3}{4^2}\right)\)

\(= \ln\left(\frac{25}{2}\right)\)