Let \(f(x) = \frac{4x^2 + 7x + 4}{(2x + 1)(x + 2)}\).
(i) Express \(f(x)\) in partial fractions.
(ii) Show that \(\int_0^4 f(x) \, dx = 8 - \ln 3\).
Solution
(i) Express \(f(x)\) in the form \(A + \frac{B}{2x+1} + \frac{C}{x+2}\).
Equating coefficients, we find \(A = 2\), \(B = 1\), and \(C = -2\).
Thus, \(f(x) = 2 + \frac{1}{2x+1} - \frac{2}{x+2}\).
(ii) Integrate \(f(x)\):
\(\int f(x) \, dx = \int \left( 2 + \frac{1}{2x+1} - \frac{2}{x+2} \right) \, dx\)
= \(2x + \frac{1}{2} \ln |2x+1| - 2 \ln |x+2| + C\).
Evaluate from 0 to 4:
\(\left[ 2x + \frac{1}{2} \ln |2x+1| - 2 \ln |x+2| \right]_0^4\)
= \(\left( 8 + \frac{1}{2} \ln 9 - 2 \ln 6 \right) - \left( 0 + \frac{1}{2} \ln 1 - 2 \ln 2 \right)\)
= \(8 + \frac{1}{2} \ln 9 - 2 \ln 6 + 2 \ln 2\)
= \(8 + \frac{1}{2} \ln 9 - 2 \ln 3\)
= \(8 - \ln 3\).
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