Ta có $\frac{(x-2)lnx+x}{x(1+lnx)}=\dfrac{x(1+\ln x) -2\ln x}{x(1+\ln x)}=1-\dfrac{2\ln x}{x(1+\ln x)}$
Vậy $I=\int_1^e \bigg (1-\dfrac{2\ln x}{x(1+\ln x)}\bigg )dx=x\bigg |_1^e -2\int_1^e \dfrac{\ln x}{\ln x +1}d(\ln x)$
$=e-1 -2\int_1^e \bigg (1 -\dfrac{1}{\ln x +1}\bigg )d(\ln x)=e-1- 2\bigg (\ln x -\ln \bigg | \ln x +1 \bigg | \bigg ) \bigg |_1^e$
$e-1-2(1 -\ln 2)=e-3+2\ln 2$