(i) Express \(f(x)\) in partial fractions:
Assume \(f(x) = \frac{A}{2x+1} + \frac{B}{2x+3} + \frac{C}{(2x+3)^2}\).
Multiply through by \((2x+1)(2x+3)^2\) to clear the denominators:
\(10x + 9 = A(2x+3)^2 + B(2x+1)(2x+3) + C(2x+1)\).
Expand and equate coefficients to solve for \(A, B, C\):
\(A = 1, B = -1, C = 3\).
(ii) Integrate \(f(x)\) from 0 to 1:
\(\int_0^1 \left( \frac{1}{2x+1} - \frac{1}{2x+3} + \frac{3}{(2x+3)^2} \right) \, dx\).
Integrate each term separately:
\(\int \frac{1}{2x+1} \, dx = \frac{1}{2} \ln |2x+1|\),
\(\int \frac{1}{2x+3} \, dx = \frac{1}{2} \ln |2x+3|\),
\(\int \frac{3}{(2x+3)^2} \, dx = -\frac{3}{2(2x+3)}\).
Evaluate from 0 to 1:
\(\left[ \frac{1}{2} \ln |2x+1| - \frac{1}{2} \ln |2x+3| - \frac{3}{2(2x+3)} \right]_0^1\).
Substitute limits:
\(\frac{1}{2} \ln 3 - \frac{1}{2} \ln 5 - \frac{3}{10} + \left( -\frac{1}{2} \ln 1 + \frac{1}{2} \ln 3 + \frac{3}{6} \right)\).
Simplify to get:
\(\frac{1}{2} \ln \frac{9}{5} + \frac{1}{5}\).
