$\int_{1}^{\mathrm{e}^{2}} \frac{\ln x}{\sqrt{x}} \mathrm{~d} x=$
$\int_{1}^{\mathrm{e}^{2}} \frac{\ln x}{\sqrt{x}} \mathrm{~d} x=2 \int_{1}^{\mathrm{e}^{2}} \ln x \mathrm{~d} \sqrt{x}=2\left[\left.\sqrt{x} \ln x\right|_{1} ^{\mathrm{e}^{2}}-\int_{1}^{\mathrm{e}^{2}} \sqrt{x} \mathrm{~d}(\ln x)\right]$
$=2\left[\left.\sqrt{x} \ln x\right|_{1} ^{\mathrm{e}^{2}}-\int_{1}^{\mathrm{e}^{2}} \sqrt{x} \mathrm{~d}(\ln x)\right]=2\left(\sqrt{\mathrm{e}^{2}} \ln \mathrm{e}^{2}-\int_{1}^{\mathrm{e}^{2}} \frac{1}{\sqrt{x}} \mathrm{~d} x\right)$
$=2\left[\left.\sqrt{x} \ln x\right|_{1} ^{\mathrm{e}^{2}}-\int_{1}^{\mathrm{e}^{2}} \sqrt{x} \mathrm{~d}(\ln x)\right]=2\left(2 \mathrm{e}-\left.2 \sqrt{x}\right|_{1} ^{\mathrm{e}^{2}}\right)=4 \quad
因此答案为：4