(i) Express \(f(x)\) as a sum of partial fractions:
\(\frac{x^2 + 3x + 3}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}\)
Multiply through by \((x+1)(x+3)\):
\(x^2 + 3x + 3 = A(x+3) + B(x+1)\)
Expand and collect terms:
\(x^2 + 3x + 3 = (A+B)x + (3A + B)\)
Equate coefficients:
\(A + B = 1\)
\(3A + B = 3\)
Solve these equations to find \(A = 1\), \(B = \frac{1}{2}\), \(C = -\frac{3}{2}\).
Thus, \(f(x) = 1 + \frac{1}{2(x+1)} - \frac{3}{2(x+3)}\).
(ii) Integrate \(f(x)\):
\(\int f(x) \, dx = \int \left( 1 + \frac{1}{2(x+1)} - \frac{3}{2(x+3)} \right) \, dx\)
\(= x + \frac{1}{2} \ln|x+1| - \frac{3}{2} \ln|x+3| + C\)
Evaluate from 0 to 3:
\(\left[ x + \frac{1}{2} \ln|x+1| - \frac{3}{2} \ln|x+3| \right]_0^3\)
\(= \left( 3 + \frac{1}{2} \ln 4 - \frac{3}{2} \ln 6 \right) - \left( 0 + \frac{1}{2} \ln 1 - \frac{3}{2} \ln 3 \right)\)
\(= 3 + \frac{1}{2} \ln 4 - \frac{3}{2} \ln 6 + \frac{3}{2} \ln 3\)
\(= 3 + \frac{1}{2} \ln \left( \frac{4 \times 3}{6} \right)\)
\(= 3 + \frac{1}{2} \ln 2\)
Thus, \(\int_0^3 f(x) \, dx = 3 - \frac{1}{2} \ln 2\).
