(a) Express \(f(x)\) in partial fractions:
Assume \(\frac{15 - 6x}{(1 + 2x)(4 - x)} = \frac{A}{1 + 2x} + \frac{B}{4 - x}\).
Multiply through by \((1 + 2x)(4 - x)\) to clear the denominators:
\(15 - 6x = A(4 - x) + B(1 + 2x)\).
Expand and collect terms:
\(15 - 6x = (4A + B) + (-A + 2B)x\).
Equate coefficients:
\(4A + B = 15\)
\(-A + 2B = -6\)
Solving these equations gives \(A = 4\) and \(B = -1\).
Thus, \(f(x) = \frac{4}{1 + 2x} - \frac{1}{4 - x}\).
(b) Integrate \(f(x)\) from 1 to 2:
\(\int_1^2 \left( \frac{4}{1 + 2x} - \frac{1}{4 - x} \right) \, dx = 4 \int_1^2 \frac{1}{1 + 2x} \, dx - \int_1^2 \frac{1}{4 - x} \, dx\).
Integrate each term:
\(4 \int \frac{1}{1 + 2x} \, dx = 2 \ln |1 + 2x|\)
\(\int \frac{1}{4 - x} \, dx = -\ln |4 - x|\)
Evaluate from 1 to 2:
\(2 \ln |1 + 2(2)| - 2 \ln |1 + 2(1)| - (\ln |4 - 2| - \ln |4 - 1|)\)
\(= 2 \ln 5 - 2 \ln 3 - (\ln 2 - \ln 3)\)
\(= 2 \ln 5 - 2 \ln 3 - \ln 2 + \ln 3\)
\(= \ln \left( \frac{5^2}{3^2 \cdot 2} \right)\)
\(= \ln \left( \frac{25}{18} \right)\)
However, the mark scheme gives the final answer as \(\ln \left( \frac{50}{27} \right)\).
