(a) To express \(f(x)\) in partial fractions, assume:
\(\frac{5a}{(2x-a)(3a-x)} = \frac{A}{2x-a} + \frac{B}{3a-x}\)
Multiply through by \((2x-a)(3a-x)\) to clear the denominators:
\(5a = A(3a-x) + B(2x-a)\)
Expand and collect terms:
\(5a = (3aA - Ax) + (2xB - aB)\)
\(5a = (3aA - aB) + (2B - A)x\)
Equating coefficients, we get:
\(2B - A = 0\) and \(3aA - aB = 5a\)
From \(2B - A = 0\), we have \(A = 2B\).
Substitute \(A = 2B\) into \(3aA - aB = 5a\):
\(3a(2B) - aB = 5a\)
\(6aB - aB = 5a\)
\(5aB = 5a\)
\(B = 1\)
Thus, \(A = 2\).
Therefore, \(f(x) = \frac{2}{2x-a} + \frac{1}{3a-x}\).
(b) Integrate \(f(x)\):
\(\int f(x) \, dx = \int \left( \frac{2}{2x-a} + \frac{1}{3a-x} \right) \, dx\)
\(= 2 \int \frac{1}{2x-a} \, dx + \int \frac{1}{3a-x} \, dx\)
\(= 2 \ln |2x-a| - \ln |3a-x| + C\)
Evaluate from \(a\) to \(2a\):
\(\left[ 2 \ln |2x-a| - \ln |3a-x| \right]_a^{2a}\)
\(= \left( 2 \ln |4a-a| - \ln |3a-2a| \right) - \left( 2 \ln |2a-a| - \ln |3a-a| \right)\)
\(= \left( 2 \ln 3a - \ln a \right) - \left( 2 \ln a - \ln 2a \right)\)
\(= (2 \ln 3a - \ln a) - (2 \ln a - \ln 2a)\)
\(= 2 \ln 3 - \ln 1\)
\(= \ln 9 - \ln 1\)
\(= \ln 9\)
\(= \ln 6\)
