Đặt \(\sqrt{1-lnx}=u\Rightarrow lnx=1-u^2\Rightarrow\dfrac{dx}{x}=-2u.du\)
\(\left\{{}\begin{matrix}x=1\Rightarrow u=1\\x=e\Rightarrow u=0\end{matrix}\right.\)
\(I=\int\limits^0_1u.\left(-2u\right)du=\int\limits^1_02u^2du=\dfrac{2}{3}u^3|^1_0=\dfrac{2}{3}\)