First, express the integrand \(\frac{u-1}{u+1}\) in the form \(A + \frac{B}{u+1}\).

We have:

\(\frac{u-1}{u+1} = 1 - \frac{2}{u+1}\)

Thus, \(A = 1\) and \(B = -2\).

Now, integrate:

\(\int_1^2 \left( 1 - \frac{2}{u+1} \right) \, du = \int_1^2 1 \, du - 2 \int_1^2 \frac{1}{u+1} \, du\)

The first integral is:

\(\int_1^2 1 \, du = [u]_1^2 = 2 - 1 = 1\)

The second integral is:

\(-2 \int_1^2 \frac{1}{u+1} \, du = -2 [\ln|u+1|]_1^2\)

Evaluate the limits:

\(-2 (\ln(3) - \ln(2)) = -2 \ln \frac{3}{2}\)

Combine the results:

\(1 - 2 \ln \frac{3}{2} = 1 + \ln \left( \frac{1}{\left(\frac{3}{2}\right)^2} \right) = 1 + \ln \frac{4}{9}\)