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