(i) Express \(f(x)\) as partial fractions:
Assume \(f(x) = \frac{A}{2x+1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}\).
Multiply through by \((2x+1)(x+1)^2\) to clear the denominators:
\(7x + 4 = A(x+1)^2 + B(2x+1)(x+1) + C(2x+1)\).
Expand and equate coefficients to solve for \(A, B, C\):
\(A = 2, \; B = -1, \; C = 3\).
Thus, \(f(x) = \frac{2}{2x+1} - \frac{1}{x+1} + \frac{3}{(x+1)^2}\).
(ii) Integrate \(f(x)\) from 0 to 2:
\(\int_0^2 \left( \frac{2}{2x+1} - \frac{1}{x+1} + \frac{3}{(x+1)^2} \right) \, dx\).
Integrate each term separately:
\(\int \frac{2}{2x+1} \, dx = \ln |2x+1|\),
\(\int \frac{-1}{x+1} \, dx = -\ln |x+1|\),
\(\int \frac{3}{(x+1)^2} \, dx = -\frac{3}{x+1}\).
Evaluate from 0 to 2:
\(\left[ \ln |2x+1| - \ln |x+1| - \frac{3}{x+1} \right]_0^2\).
Calculate the definite integral:
At \(x = 2\): \(\ln 5 - \ln 3 - \frac{3}{3} = \ln \frac{5}{3} - 1\).
At \(x = 0\): \(\ln 1 - \ln 1 - 3 = -3\).
Result: \((\ln \frac{5}{3} - 1) - (-3) = \ln \frac{5}{3} + 2\).
Thus, \(\int_0^2 f(x) \, dx = 2 + \ln \frac{5}{3}\).
