To solve \(\int_{1}^{4} \frac{\ln x}{\sqrt{x}} \, dx\), we use integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{\sqrt{x}} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = 2\sqrt{x}\).

Applying integration by parts:

\(\int u \, dv = uv - \int v \, du\)

\(\int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x} \ln x \bigg|_{1}^{4} - \int 2\sqrt{x} \cdot \frac{1}{x} \, dx\)

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 2 \int x^{-1/2} \, dx\)

Integrate \(\int x^{-1/2} \, dx\):

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 2 \cdot 2x^{1/2} \bigg|_{1}^{4}\)

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 4\sqrt{x} \bigg|_{1}^{4}\)

Evaluate at the limits:

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

\(= (4 \ln 4 - 8) - (0 - 4)\)

\(= 4 \ln 4 - 8 + 4\)

\(= 4(\ln 4 - 1)\)