To solve \(\int_{1}^{2} x \ln x \, dx\), we use integration by parts. Let \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{1}{2} x^2\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

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

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

\(= \frac{1}{2} x^2 \ln x - \frac{1}{2} \cdot \frac{1}{2} x^2 + C\)

\(= \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 + C\)

Now, evaluate from 1 to 2:

\(\left[ \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 \right]_{1}^{2}\)

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

\(= \left( 2 \ln 2 - 1 \right) - \left( 0 - \frac{1}{4} \right)\)

\(= 2 \ln 2 - 1 + \frac{1}{4}\)

\(= 2 \ln 2 - \frac{3}{4}\)