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 \(v = \int x^{-\frac{3}{2}} \, dx = -2x^{-\frac{1}{2}}\).
Applying integration by parts:
\(\int u \, dv = uv - \int v \, du\)
\(= \left[ -2x^{-\frac{1}{2}} \ln x \right]_{1}^{4} + 2 \int_{1}^{4} x^{-\frac{1}{2}} \, dx\)
Calculate \(\int x^{-\frac{1}{2}} \, dx\):
\(\int x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}}\)
Substitute back:
\(= \left[ -2x^{-\frac{1}{2}} \ln x \right]_{1}^{4} + 2 \left[ 2x^{\frac{1}{2}} \right]_{1}^{4}\)
Evaluate at the bounds:
\(= \left[ -2(4)^{-\frac{1}{2}} \ln 4 + 4(4)^{\frac{1}{2}} \right] - \left[ -2(1)^{-\frac{1}{2}} \ln 1 + 4(1)^{\frac{1}{2}} \right]\)
\(= \left[ -\ln 4 + 8 \right] - \left[ 0 + 4 \right]\)
\(= -\ln 4 + 8 - 4\)
\(= 4 - \ln 4\)
Thus, the integral evaluates to \(2 - \ln 4\).
