(a) Express \(f(x)\) in partial fractions:
Assume \(f(x) = \frac{A}{2x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}\).
Multiply through by \((2x+1)(x+2)^2\) to clear the denominators:
\(3 - 3x^2 = A(x+2)^2 + B(2x+1)(x+2) + C(2x+1)\).
Expand and equate coefficients to find \(A = 1\), \(B = -2\), \(C = 3\).
Thus, \(f(x) = \frac{1}{2x+1} - \frac{2}{x+2} + \frac{3}{(x+2)^2}\).
(b) Integrate \(f(x)\) from 0 to 4:
\(\int_0^4 f(x) \, dx = \int_0^4 \left( \frac{1}{2x+1} - \frac{2}{x+2} + \frac{3}{(x+2)^2} \right) \, dx\).
Integrate each term separately:
\(\int \frac{1}{2x+1} \, dx = \frac{1}{2} \ln |2x+1|\),
\(\int \frac{-2}{x+2} \, dx = -2 \ln |x+2|\),
\(\int \frac{3}{(x+2)^2} \, dx = \frac{-3}{x+2}\).
Evaluate from 0 to 4:
\(\left[ \frac{1}{2} \ln |2x+1| - 2 \ln |x+2| - \frac{3}{x+2} \right]_0^4\).
Substitute limits:
\(\left( \frac{1}{2} \ln 9 - 2 \ln 6 - \frac{3}{6} \right) - \left( \frac{1}{2} \ln 1 - 2 \ln 2 - \frac{3}{2} \right)\).
Simplify to obtain \(1 - \ln 3\).
