$$f(\alpha ) = \int_1^\alpha {{{{{\log }_{10}}t} \over {1 + t}}dt} $$ ...... (i)

$$f\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt} $$

Substituting $$t \to {1 \over p}$$

$$f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}\left( {{1 \over p}} \right)} \over {1 + {1 \over p}}}\left( {{{ - 1} \over {{p^2}}}} \right)dp} $$

$$ = \int_1^\alpha {{{{{\log }_{10}}p} \over {p(p + 1)}}dp = \int_1^\alpha {\left( {{{{{\log }_{10}}t} \over t} - {{{{\log }_{10}}t} \over {t + 1}}} \right)dt} } $$ ....... (ii)

By (i) + (ii)

$$f(\alpha ) + f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}t} \over t}dt = \int_1^\alpha {{{\ln t} \over t}\,.\,{{\log }_{10}}e\,dt} } $$

$$ = {{{{(\ln \alpha )}^2}} \over {2{{\log }_e}10}}$$

$$\alpha = {e^3} \Rightarrow f({e^3}) + f({e^{ - 3}}) = {9 \over {2{{\log }_e}10}}$$