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