Explanation

(b)
Let $I = \int_{e^{- 1}}^{e^{2}}{\left| \frac{\log_{e}x}{x} \right|dx}$
$= \int_{e^{- 1}}^{1}{\left| \frac{\log_{e}x}{x} \right|dx} + \int_{1}^{e^{2}}{\left| \frac{\log_{e}x}{x} \right|dx}$
$= \int_{e^{- 1}}^{1}{\left( - \frac{\log_{e}x}{x} \right)dx} + \int_{1}^{e^{2}}{\frac{\log_{e}x}{x}dx}$
Put $\log_{e}{x = \mathcal{z\ \ \Rightarrow \ }\frac{1}{x}dx = d\mathcal{z}}$
$\therefore\ I = \int_{- 1}^{0}{\left( \mathcal{- z} \right)d\mathcal{z +}\int_{0}^{2}{\mathcal{z\ }d\mathcal{z}}}$
$= \left\lbrack - \frac{\mathcal{z}^{2}}{2} \right\rbrack_{- 1}^{0} + \left\lbrack \frac{\mathcal{z}^{2}}{2} \right\rbrack_{0}^{2}$
$= - \frac{1}{2}\lbrack 0 - 1\rbrack + \frac{1}{2}\lbrack 4 - 0\rbrack$
$= \frac{1}{2} + 2 = \frac{5}{2}$