(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{x^2 + 9x}{(3x - 1)(x^2 + 3)} = \frac{A}{3x - 1} + \frac{Bx + C}{x^2 + 3}\).

Multiply through by \((3x - 1)(x^2 + 3)\) to clear the denominators:

\(x^2 + 9x = A(x^2 + 3) + (Bx + C)(3x - 1)\).

Expand and collect like terms:

\(x^2 + 9x = Ax^2 + 3A + 3Bx^2 - Bx + Cx - C\).

Combine terms:

\(x^2 + 9x = (A + 3B)x^2 + (-B + C)x + (3A - C)\).

Equate coefficients:

\(A + 3B = 1\)

\(-B + C = 9\)

\(3A - C = 0\)

Solving these equations gives \(A = 1\), \(B = 0\), \(C = 3\).

Thus, \(f(x) = \frac{1}{3x - 1} + \frac{3}{x^2 + 3}\).

(b) Integrate \(f(x)\) from 1 to 3:

\(\int_1^3 \left( \frac{1}{3x - 1} + \frac{3}{x^2 + 3} \right) \, dx = \int_1^3 \frac{1}{3x - 1} \, dx + \int_1^3 \frac{3}{x^2 + 3} \, dx\).

Integrate each term separately:

\(\int \frac{1}{3x - 1} \, dx = \frac{1}{3} \ln |3x - 1|\).

\(\int \frac{3}{x^2 + 3} \, dx = \sqrt{3} \arctan \left( \frac{x}{\sqrt{3}} \right)\).

Evaluate from 1 to 3:

\(\left[ \frac{1}{3} \ln |3x - 1| \right]_1^3 = \frac{1}{3} \ln 8 - \frac{1}{3} \ln 2\).

\(\left[ \sqrt{3} \arctan \left( \frac{x}{\sqrt{3}} \right) \right]_1^3 = \sqrt{3} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\sqrt{3} \pi}{6}\).

Combine results:

\(\frac{1}{3} \ln 8 - \frac{1}{3} \ln 2 + \frac{\sqrt{3} \pi}{6} = \frac{2}{3} \ln 2 + \frac{\sqrt{3} \pi}{6}\).