To solve \(\int_{1}^{4} x^{\frac{3}{2}} \ln x \, dx\), we use integration by parts. Let \(u = \ln x\) and \(dv = x^{\frac{3}{2}} \, dx\).
Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{2}{5} x^{\frac{5}{2}}\).
Applying integration by parts:
\(\int u \, dv = uv - \int v \, du\)
\(= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \int x^{\frac{5}{2}} \cdot \frac{1}{x} \, dx\)
\(= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \int x^{\frac{3}{2}} \, dx\)
Integrate \(\int x^{\frac{3}{2}} \, dx\):
\(= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \cdot \frac{2}{5} x^{\frac{5}{2}}\)
\(= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{4}{25} x^{\frac{5}{2}}\)
Evaluate from 1 to 4:
\(\left[ \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{4}{25} x^{\frac{5}{2}} \right]_{1}^{4}\)
\(= \left( \frac{2}{5} \cdot 32 \ln 4 - \frac{4}{25} \cdot 32 \right) - \left( \frac{2}{5} \cdot 1 \cdot 0 - \frac{4}{25} \cdot 1 \right)\)
\(= \frac{64}{5} \ln 4 - \frac{128}{25} + \frac{4}{25}\)
\(= \frac{64}{5} \ln 4 - \frac{124}{25}\)
Since \(\ln 4 = 2 \ln 2\), substitute:
\(= \frac{128}{5} \ln 2 - \frac{124}{25}\)
