$$\int_1^3 {\left| {\left( {2 - x} \right){{\log }_e}x} \right|dx} $$     is equal to :

Solution :
$$\eqalign{ & \int_1^3 {\left| {2 - x} \right|\,\left| {{{\log }_e}x} \right|dx} = \int_1^2 {\left| {2 - x} \right|\,\left| {{{\log }_e}x} \right|dx} + \int_2^3 {\left| {2 - x} \right|\,\left| {{{\log }_e}x} \right|dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_1^2 {\left( {2 - x} \right){{\log }_e}x\,dx} + \int_2^3 { - \left( {2 - x} \right){{\log }_e}x\,dx} \cr & {\text{Now, }}\int {\left( {2 - x} \right)} {\log _e}x\,dx = {\log _e}x.\left( {2x - \frac{{{x^2}}}{2}} \right) - \int {\left( {2x - \frac{{{x^2}}}{2}} \right).\frac{1}{x}dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2x - \frac{{{x^2}}}{2}} \right){\log _e}x - \int {\left( {2 - \frac{x}{2}} \right)dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2x - \frac{{{x^2}}}{2}} \right){\log _e}x - \left( {2x - \frac{{{x^2}}}{4}} \right) \cr & \therefore I = \left[ {\left( {2x - \frac{{{x^2}}}{2}} \right){{\log }_e}x - \left( {2x - \frac{{{x^2}}}{4}} \right)} \right]_1^2 + \left[ { - \left( {2x - \frac{{{x^2}}}{2}} \right){{\log }_e}x + \left( {2x - \frac{{{x^2}}}{4}} \right)} \right]_2^3 \cr & = \left( {4 - 2} \right){\log _e}2 - \left( {4 - 1} \right) + \left( {2 - \frac{1}{4}} \right) + \left\{ { - \left( {6 - \frac{9}{2}} \right){{\log }_e}3 + \left( {6 - \frac{9}{4}} \right) + \left( {4 - 2} \right){{\log }_e}2 - \left( {4 - 1} \right)} \right\} \cr & = 2{\log _e}2 - 3 + \frac{7}{4} - \frac{3}{2}{\log _e}3 + \frac{{15}}{4} + 2{\log _e}2 - 3 \cr & = 4{\log _e}2 - \frac{3}{2}{\log _e}3 - \frac{1}{2} \cr & = \log \frac{{16}}{{3\sqrt 3 }} - \frac{1}{2} \cr} $$