Đặt [TEX]I=\int_{e}^{e^5}\frac{lnxdx}{\sqrt{2lnx-1}}[/TEX]
[TEX]2I=\int_{e}^{e^5}\frac{2lnxdx}{\sqrt{2lnx-1}}=\int_{e}^{e^5}[\sqrt{2lnx-1}+\frac{1}{\sqrt{2lnx-1}}]dx[/TEX]
Ta có [TEX]\int_{e}^{e^5}\sqrt{2lnx-1}dx=x.\sqrt{2lnx-1}|_e^{e^5}-\int_{e}^{e^5}\frac{dx}{\sqrt{2lnx-1}}[/TEX]