(i) Let \(x = t^2 + 1\). Then \(dx = 2t \, dt\).

When \(t = 0\), \(x = 0^2 + 1 = 1\).

When \(t = 2\), \(x = 2^2 + 1 = 5\).

Substitute into the integral:

\(I = \int_0^2 4t^3 \ln(t^2 + 1) \, dt = \int_1^5 (2x - 2) \ln x \, dx\).

(ii) Use integration by parts:

Let \(u = \ln x\) and \(dv = (2x - 2) \, dx\).

Then \(du = \frac{1}{x} \, dx\) and \(v = x^2 - 2x\).

\(\int (2x - 2) \ln x \, dx = (x^2 - 2x) \ln x - \int (x^2 - 2x) \frac{1}{x} \, dx\).

\(= (x^2 - 2x) \ln x - \int (x - 2) \, dx\).

\(= (x^2 - 2x) \ln x - \left( \frac{1}{2}x^2 - 2x \right)\).

Evaluate from 1 to 5:

\(= [(5^2 - 2 \cdot 5) \ln 5 - \left( \frac{1}{2} \cdot 5^2 - 2 \cdot 5 \right)] - [(1^2 - 2 \cdot 1) \ln 1 - \left( \frac{1}{2} \cdot 1^2 - 2 \cdot 1 \right)]\).

\(= [15 \ln 5 - (\frac{25}{2} - 10)] - [0 - (\frac{1}{2} - 2)]\).

\(= 15 \ln 5 - \frac{5}{2} + \frac{3}{2}\).

\(= 15 \ln 5 - 4\).