Explanation

(b)
Let $I = \int_{1}^{e}{10^{\log_{e}x}dx}$
Again, let $I_{1} = \int_{}^{}10^{\log_{e}x}\ dx$
$\Rightarrow I_{1} = x.10^{\log_{e}x} - \int_{}^{}{x.10^{\log_{e}{x.}}\frac{\log_{e}10}{x}dx}$
$\Rightarrow I_{1} = x.10^{\log_{e}x} - \int_{}^{}{10^{\log_{e}{x\ \ }}\log_{e}{10dx}}\$
$\Rightarrow (1 + \log_{e}{10)I_{1} = x.10^{\log_{e}x}}$
$\Rightarrow I_{1} = \frac{x.10^{\log_{e}x}}{1 + \log_{e}10}$
$\therefore I = \left\lbrack \frac{x.10^{\log_{e}x}}{1 + \log_{e}10} \right\rbrack_{1}^{e}$
$= \left\lbrack \frac{10_{e} - 1}{1 + \log_{e}10} \right\rbrack = \frac{10e - 1}{\log_{e}{10e}}$