(i) First, find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\):

\(\frac{dy}{dt} = 4 + \frac{2}{2t - 1}\)

\(\frac{dx}{dt} = 2t\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4 + \frac{2}{2t - 1}}{2t}\)

Simplify to obtain:

\(\frac{dy}{dx} = \frac{8t - 2}{2t(2t - 1)}\)

(ii) To find the equation of the normal, first find the gradient of the tangent at \(t = 1\):

Substitute \(t = 1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{8(1) - 2}{2(1)(2(1) - 1)} = \frac{6}{2} = 3\)

The gradient of the normal is the negative reciprocal of the tangent's gradient:

\(m_{\text{normal}} = -\frac{1}{3}\)

Find the point on the curve at \(t = 1\):

\(x = 1^2 + 1 = 2\)

\(y = 4(1) + \ln(2(1) - 1) = 4 + \ln(1) = 4\)

Using the point-slope form of the line equation:

\(y - 4 = -\frac{1}{3}(x - 2)\)

Rearrange to the form \(ax + by + c = 0\):

\(3(y - 4) = -(x - 2)\)

\(3y - 12 = -x + 2\)

\(x + 3y - 14 = 0\)