(i) Express \(\frac{2}{(x+1)(x+3)}\) as \(\frac{A}{x+1} + \frac{B}{x+3}\).

Multiply through by \((x+1)(x+3)\) to get: \(2 = A(x+3) + B(x+1)\).

Set \(x = -1\): \(2 = A(-1+3) \Rightarrow A = 1\).

Set \(x = -3\): \(2 = B(-3+1) \Rightarrow B = -1\).

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

(ii) Square the result from part (i):

\(\left( \frac{1}{x+1} - \frac{1}{x+3} \right)^2 = \frac{1}{(x+1)^2} - \frac{2}{(x+1)(x+3)} + \frac{1}{(x+3)^2}\).

Substitute \(\frac{2}{(x+1)(x+3)} = \frac{1}{x+1} - \frac{1}{x+3}\) to get:

\(\frac{1}{(x+1)^2} - \frac{1}{x+1} + \frac{1}{x+3} + \frac{1}{(x+3)^2}\).

(iii) Integrate \(\int_0^1 \frac{4}{(x+1)^2(x+3)^2} \, dx\).

Using the result from part (ii):

\(\int_0^1 \left( \frac{1}{(x+1)^2} - \frac{1}{x+1} + \frac{1}{x+3} + \frac{1}{(x+3)^2} \right) \, dx\).

Integrate each term:

\(\int \frac{1}{(x+1)^2} \, dx = -\frac{1}{x+1}\), \(\int \frac{1}{x+1} \, dx = \ln(x+1)\), \(\int \frac{1}{x+3} \, dx = \ln(x+3)\), \(\int \frac{1}{(x+3)^2} \, dx = -\frac{1}{x+3}\).

Evaluate from 0 to 1:

\(-\frac{1}{2} - \ln 2 + \ln 4 - \frac{1}{4} = \frac{7}{12} - \ln \frac{3}{2}\).