(i) Express \(f(x)\) in partial fractions:
Assume \(\frac{3x^2 - 4}{x^2(3x + 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x + 2}\).
Multiply through by \(x^2(3x + 2)\) to clear the denominators:
\(3x^2 - 4 = A(x)(3x + 2) + B(3x + 2) + C(x^2)\).
Expand and collect terms:
\(3x^2 - 4 = (3A + C)x^2 + (2A + 3B)x + 2B\).
Equate coefficients:
\(3A + C = 3\)
\(2A + 3B = 0\)
\(2B = -4\)
Solve for \(B\):
\(B = -2\).
Substitute \(B = -2\) into \(2A + 3B = 0\):
\(2A - 6 = 0\)
\(A = 3\).
Substitute \(A = 3\) into \(3A + C = 3\):
\(9 + C = 3\)
\(C = -6\).
Thus, \(f(x) = \frac{3}{x} - \frac{2}{x^2} - \frac{6}{3x+2}\).
(ii) Integrate \(f(x)\) from 1 to 2:
\(\int_1^2 \left( \frac{3}{x} - \frac{2}{x^2} - \frac{6}{3x+2} \right) \, dx\).
Integrate each term separately:
\(\int \frac{3}{x} \, dx = 3 \ln |x|\)
\(\int \frac{-2}{x^2} \, dx = \frac{2}{x}\)
\(\int \frac{-6}{3x+2} \, dx = -2 \ln |3x+2|\)
Evaluate from 1 to 2:
\(\left[ 3 \ln x + \frac{2}{x} - 2 \ln(3x+2) \right]_1^2\)
\(= \left( 3 \ln 2 + 1 - 2 \ln 8 \right) - \left( 0 + 2 - 2 \ln 5 \right)\)
\(= 3 \ln 2 - 2 \ln 8 - 2 + 2 \ln 5\)
\(= \ln\left(\frac{8}{8}\right) - 1\)
\(= \ln\left(\frac{25}{8}\right) - 1\).
