\(I=\int_{1}^{e}\frac{1}{x}dx-2\int_{1}^{e}\frac{lnx}{x^2}dx=ln\left | x \right | \bigg |^e_1-2\int_{1}^{e}\frac{lnx}{x^2}dx=1-2\int_{1}^{e}\frac{lnx}{x^2}dx\) Tính \(J=\int_{1}^{e}\frac{lnx}{x^2}dx\) Đặt \(u=lnx,dv=\frac{1}{x^2}dx\). Khi đó \(du=\frac{1}{x}dx,v=-\frac{1}{x}\) Do đó \(J=-\frac{1}{x}lnx\bigg |^e_1+\int_{1}^{e}\frac{1}{x^2}dx\) \(J=-\frac{1}{e}-\frac{1}{x}\bigg |^e_1=-\frac{2}{e}+1\) Vậy \(I=-1+\frac{4}{e}\)