Chọn phương án D.

Đặt \(u=\sqrt{2+\ln x}\Leftrightarrow u^2=2+\ln x\).
Vi phân hai vế ta được \(2u\mathrm{d}u=\dfrac{1}{x}\mathrm{d}x\).

• \(x=1\Rightarrow u=\sqrt{2}\)
• \(x=\mathrm{e}\Rightarrow u=\sqrt{3}\)

\(\begin{aligned}
I&=\displaystyle\int\limits_{1}^{\mathrm{e}}\dfrac{\sqrt{2+\ln x}}{2x}\mathrm{\,d}x=\displaystyle\int\limits_{\sqrt{2}}^{\sqrt{3}}\dfrac{u}{2}\cdot2u\mathrm{\,d}u\\
&=\displaystyle\int\limits_{\sqrt{2}}^{\sqrt{3}}u^2\mathrm{\,d}u=\dfrac{u^3}{3}\bigg|_{\sqrt{2}}^{\sqrt{3}}\\
&=\dfrac{\sqrt{3}^3-\sqrt{2}^3}{3}=\dfrac{3\sqrt{3}-2\sqrt{2}}{3}.
\end{aligned}\)