(i) To express \(\frac{1}{x^2(2x+1)}\) in the form \(\frac{A}{x^2} + \frac{B}{x} + \frac{C}{2x+1}\), we equate:

\(\frac{1}{x^2(2x+1)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C}{2x+1}\)

Multiply through by \(x^2(2x+1)\) to clear the denominators:

\(1 = A(2x+1) + Bx(2x+1) + Cx^2\)

Expanding and collecting terms gives:

\(1 = (2A)x + A + (2B)x^2 + Bx + Cx^2\)

Equating coefficients, we get:

\(A = 1\), \(2B + C = 0\), \(2A + B = 0\).

Solving these, we find \(A = 1\), \(B = -2\), \(C = 4\).

(ii) Given \(y = x^2(2x+1) \frac{dy}{dx}\), separate variables:

\(\frac{1}{y} dy = x^2(2x+1) dx\)

Integrate both sides:

\(\ln y = -\frac{1}{2} \ln x + 2 \ln (2x+1) + c\)

Using the initial condition \(y = 1\) when \(x = 1\), solve for \(c\):

\(\ln 1 = -\frac{1}{2} \ln 1 + 2 \ln 3 + c\)

\(c = -2 \ln 3\)

Thus, \(\ln y = -\frac{1}{2} \ln x + 2 \ln (2x+1) - 2 \ln 3\).

Exponentiate to solve for \(y\):

\(y = \frac{(2x+1)^2}{3^2 x^{1/2}}\)

Substitute \(x = 2\):

\(y = \frac{(5)^2}{9 \sqrt{2}} = \frac{25}{36} e^{\frac{1}{2}}\)