\(I=\int_{1}^{e}\left ( \frac{\sqrt{3+lnx}}{x} +2lnx\right )dx=\int_{1}^{e}\frac{\sqrt{3+lnx}}{x}dx+\int_{1}^{e}2lnxdx=J+K\) Ta có \(K=\int_{1}^{e}2lnxdx=2xlnx |_{1}^{e}-\int_{1}^{e}2dx=2xlnx|_{1}^{e}-2x|_{1}^{e}=2\) Đặt \(t=\sqrt{3+lnx}\Rightarrow t^2=3+lnx\Rightarrow 2tdt=\frac{dx}{x}\) Khi đó \(J=\int_{\sqrt{3}}^{2}2t^2dt=\frac{2}{3}t^3 |_{\sqrt{3}}^{2}=\frac{16-6\sqrt{3}}{3}\)