The value of the integral $$\int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\frac{{{{\log }_e}x}}{x}} \right|dx} ,$$   is:

Solution :
$${\text{Let }}I = \int_{{e^{ - 1}}}^{{e^2}} {\left| {\frac{{{{\log }_e}x}}{x}} \right|dx} $$
We know that for $$\frac{1}{e} < x < 1,\,{\log _e}x < 0$$     and hence $$\frac{{{{\log }_e}x}}{x} < 0$$
and for $$1 < x < {e^2},\log \,x > 0$$     and hence $$\frac{{{{\log }_e}x}}{x} > 0$$
$$\eqalign{ & \therefore I = \int_{\frac{1}{e}}^1 {\left[ { - \frac{{{{\log }_e}\,x}}{x}} \right]dx + } \int_1^{{e^2}} {\frac{{{{\log }_e}\,x}}{x}dx} \cr & = - \frac{1}{2}\left[ {{{\left( {{{\log }_e}\,x} \right)}^2}} \right]_{\frac{1}{e}}^1 + \frac{1}{2}\left[ {{{\left( {{{\log }_e}\,x} \right)}^2}} \right]_1^{{e^2}} \cr & = \frac{1}{2} + 2 \cr & = \frac{5}{2} \cr} $$