To solve \(\int_1^2 \frac{2}{u(4-u)} \, du\), we use partial fraction decomposition.

Express \(\frac{2}{u(4-u)}\) as \(\frac{A}{u} + \frac{B}{4-u}\).

Equating coefficients, we have:

\(2 = A(4-u) + Bu\)

\(2 = 4A - Au + Bu\)

\(2 = 4A + (B-A)u\)

Comparing coefficients, we get:

\(4A = 2\) and \(B-A = 0\)

Solving these, \(A = \frac{1}{2}\) and \(B = \frac{1}{2}\).

Thus, \(\frac{2}{u(4-u)} = \frac{1/2}{u} + \frac{1/2}{4-u}\).

Integrate each term separately:

\(\int \frac{1/2}{u} \, du = \frac{1}{2} \ln |u|\)

\(\int \frac{1/2}{4-u} \, du = -\frac{1}{2} \ln |4-u|\)

Evaluate from 1 to 2:

\(\left[ \frac{1}{2} \ln |u| - \frac{1}{2} \ln |4-u| \right]_1^2\)

\(= \left( \frac{1}{2} \ln 2 - \frac{1}{2} \ln 2 \right) - \left( \frac{1}{2} \ln 1 - \frac{1}{2} \ln 3 \right)\)

\(= 0 - (-\frac{1}{2} \ln 3)\)

\(= \frac{1}{2} \ln 3\)