(i) Express \(f(x)\) as:
\(f(x) = \frac{A}{2x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}\)
Multiply through by \((2x-1)(x+2)^2\) to clear the denominators:
\(11x + 7 = A(x+2)^2 + B(2x-1)(x+2) + C(2x-1)\)
Expand and equate coefficients to find \(A = 2\), \(B = -1\), \(C = 3\).
Thus, \(f(x) = \frac{2}{2x-1} - \frac{1}{x+2} + \frac{3}{(x+2)^2}\).
(ii) Integrate each term separately:
\(\int \frac{2}{2x-1} \, dx = \ln|2x-1|\)
\(\int -\frac{1}{x+2} \, dx = -\ln|x+2|\)
\(\int \frac{3}{(x+2)^2} \, dx = -\frac{3}{x+2}\)
Evaluate from 1 to 2:
\(\left[ \ln|2x-1| - \ln|x+2| - \frac{3}{x+2} \right]_1^2\)
\(= \left( \ln(3) - \ln(4) - \frac{3}{4} \right) - \left( \ln(1) - \ln(3) - \frac{3}{3} \right)\)
\(= \ln\left(\frac{3}{4}\right) + 1 - \left(0 - \ln(3) - 1\right)\)
\(= \ln\left(\frac{3}{4}\right) + 1 + \ln(3) + 1\)
\(= \ln\left(\frac{9}{4}\right) + \frac{1}{4}\)
