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Ta có:
Đặt $u = \,\,\ln x\,\, \Rightarrow \,du = \frac{{dx}}{x}$
$\begin{array}{l}
x = 1\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,u = 0\\
x = \sqrt e \,\,\,\,\, \Rightarrow \,\,\,u = \frac{1}{2}\\
\,\,\,\Rightarrow I = \int\limits_0^{\frac{1}{2}} {\frac{{du}}{{\sqrt {1 - {u^2}} }} =[\left. {arc\sin \,u}
\right]_0^{\frac{1}{2}}} = arc\sin \frac{1}{2} = \frac{\pi }{6}
\end{array}$
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