(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:

\(\frac{1}{2} \ln \frac{9}{5} + \frac{1}{5}\).