(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{2}{(2x-1)(2x+1)} = \frac{A}{2x-1} + \frac{B}{2x+1}\).

Multiply through by \((2x-1)(2x+1)\) to get:

\(2 = A(2x+1) + B(2x-1)\).

Expanding gives \(2 = (2A + 2B)x + (A - B)\).

Equating coefficients, we have:

\(2A + 2B = 0\) and \(A - B = 2\).

Solving these equations, we find \(A = 1\) and \(B = -1\).

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

(b) Square the result of part (a):

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

Expanding gives:

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

Using \(f(x) = \frac{1}{2x-1} - \frac{1}{2x+1}\), we have:

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

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

(c) Integrate \((f(x))^2\) from 1 to 2:

\(\int_1^2 (f(x))^2 \, dx = \int_1^2 \left( \frac{1}{(2x-1)^2} - \frac{1}{2x-1} + \frac{1}{2x+1} + \frac{1}{(2x+1)^2} \right) \, dx\).

Integrate each term separately:

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

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

\(\int \frac{1}{2x+1} \, dx = \frac{1}{2} \ln|2x+1|\),

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

Evaluate from 1 to 2:

\(\left[ -\frac{1}{2(2x-1)} - \frac{1}{2} \ln|2x-1| + \frac{1}{2} \ln|2x+1| - \frac{1}{2(2x+1)} \right]_1^2\).

Substitute limits and simplify to obtain:

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