Explanation

(c)
An anti-derivative of $f(x)$
$= F(x)$
$= \int_{}^{}{\lbrack\log{(\log{x) + \left( {\log{x)}}^{- 2} \right\rbrack dx + c}}}$
$= x\log{(\log{x) - \int_{}^{}{\frac{x}{x\log x}dx + \int_{}^{}{({\log{x)}}^{- 2}dx + c}}}}$
(integrating by parts the first integral)
$= x\log{(\log{x) - \lbrack x({\log{x)}}^{- 1} + \int_{}^{}{\left( {\log{x)}}^{- 2}dx \right\rbrack + \int_{}^{}\left( \log x \right)^{- 2}dx + c}}}$
[again integrating by parts $({\log{x)}}^{- 1}$]
$= x\log{(\log{x) - x({\log{x)}}^{- 1} + c}}$
On putting $x = e$, we have, $e = 0 - e + c$ so, $c = 2e$
Thus, $F(x) = x\lbrack\log{(\log{x) - \left( {\log{x)}}^{- 1} \right\rbrack + 2e}}$