(i) To express \(f(x)\) in the form \(A + \frac{B}{x+2} + \frac{C}{2x-1}\), perform partial fraction decomposition:
\(\frac{4x^2 + 9x - 8}{(x+2)(2x-1)} = A + \frac{B}{x+2} + \frac{C}{2x-1}\)
Multiply through by \((x+2)(2x-1)\) to clear the denominators:
\(4x^2 + 9x - 8 = A(x+2)(2x-1) + B(2x-1) + C(x+2)\)
Expand and equate coefficients to solve for \(A\), \(B\), and \(C\):
\(A = 2, B = 2, C = -1\).
(ii) Integrate \(f(x)\) using the partial fraction decomposition:
\(\int f(x) \, dx = \int \left( 2 + \frac{2}{x+2} - \frac{1}{2x-1} \right) \, dx\)
\(= \int 2 \, dx + \int \frac{2}{x+2} \, dx - \int \frac{1}{2x-1} \, dx\)
\(= 2x + 2 \ln |x+2| - \frac{1}{2} \ln |2x-1| + C\)
Evaluate from 1 to 4:
\(\left[ 2x + 2 \ln |x+2| - \frac{1}{2} \ln |2x-1| \right]_1^4\)
\(= \left( 8 + 2 \ln 6 - \frac{1}{2} \ln 7 \right) - \left( 2 + 2 \ln 3 - \frac{1}{2} \ln 1 \right)\)
\(= 6 + 2 \ln \frac{6}{3} - \frac{1}{2} \ln \frac{7}{1}\)
\(= 6 + \ln 2 - \frac{1}{2} \ln 7\)
\(= 6 + \frac{1}{2} \ln \left( \frac{16}{7} \right)\)
