令 $u = \ln x$，则当 $x = 1$ 时 $u = 0$；当 $x = e$ 时 $u = 1$；且 $du = \dfrac{1}{x}dx$，故原积分化为：$$\int_{1}^{e} \dfrac{1}{x\sqrt{\ln x}}\,dx = \int_{0}^{1} u^{-1/2}\,du = \left[2u^{1/2}\right]_{0}^{1} = 2(1 - 0) = 2.$$ 故答案为 $2$。