Explanation

(a)
Given , $\int_{2}^{e}\left( \frac{1}{\log x} - \frac{1}{({\log{x)}}^{2}} \right)dx = a + \frac{b}{\log 2}$
Put $log\ x = z \Rightarrow x = e^{z} \Rightarrow dx = e^{z}dz$
$\therefore\int_{2}^{e}{\left( \frac{1}{\log x} - \frac{1}{({\log{x)}}^{2}} \right)dx = \int_{\log 2}^{1}{\left( \frac{1}{z} - \frac{1}{z^{2}} \right)e^{z}}\ dz}$
$= \int_{\log 2}^{1}e^{2}\ \left( \frac{1}{z} + d\left( \frac{1}{z} \right) \right)dz$
$= \left\lbrack e^{2}.\frac{1}{z} \right\rbrack_{\log{2\ }}^{1} = e - \frac{2}{\log 2}$
$\therefore\ a = e\$and$\ b = - 2$