[tex]\displaystyle I = \int_{1}^e \dfrac{dx}{x\sqrt{4-\ln^2 x}}[/tex]
Facem schimbarea de variabilă:
[tex]\displaystyle \ln x = t \Rightarrow (\ln x)'\, dx = t'\, dt \Rightarrow \dfrac{1}{x}\, dx = dt\\ x = 1 \Rightarrow t = \ln 1 = 0 \\ x = e \Rightarrow t = \ln e = 1 \\ \\ I = \int_{0}^1 \dfrac{dt}{\sqrt{4-t^2}} =\int_{0}^1 \dfrac{dt}{\sqrt{2^2-t^2}} = \arcsin \dfrac{t}{2}\Bigg|_{0}^1 =\\ \\=\arcsin \dfrac{1}{2} - \arcsin 0 = \boxed{\dfrac{\pi}{6}}[/tex]
