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:
\(\left[ \frac{1}{2} \ln |u| - \frac{1}{2} \ln |4-u| \right]_1^2\)
\(= \left( \frac{1}{2} \ln 2 - \frac{1}{2} \ln 2 \right) - \left( \frac{1}{2} \ln 1 - \frac{1}{2} \ln 3 \right)\)
\(= 0 - (-\frac{1}{2} \ln 3)\)
\(= \frac{1}{2} \ln 3\)
