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

Assume \(f(x) = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+2}\).

Multiply through by the denominator \(x^2(x+2)\) to get:

\(x^2 + x + 6 = A x(x+2) + B(x+2) + C x^2\).

Equating coefficients, solve for \(A, B, C\):

\(A = -1, \; B = 3, \; C = 2\).

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

(ii) Integrate \(f(x)\) from 1 to 4:

\(\int_1^4 \left( \frac{-1}{x} + \frac{3}{x^2} + \frac{2}{x+2} \right) \, dx\).

Integrate each term separately:

\(\int \frac{-1}{x} \, dx = -\ln|x|\),

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

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

Evaluate from 1 to 4:

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

Substitute the limits:

\(\left( -\ln 4 - \frac{3}{4} + 2\ln 6 \right) - \left( -\ln 1 - 3 + 2\ln 3 \right)\).

Simplify:

\(-\ln 4 - \frac{3}{4} + 2\ln 6 + 3 - 2\ln 3\).

Combine logs:

\(-\ln 4 + 2\ln 2 + 3 - \frac{3}{4}\).

\(= \frac{9}{4}\).