If $$f(\alpha)=\int\limits_{1}^{\alpha} \frac{\log _{10} \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}, \alpha>0$$, then $f\left(\mathrm{e}^{3}\right)+f\left(\mathrm{e}^{-3}\right)$ is equal to :

<p>$f(\alpha ) = \int_1^\alpha {{{{{\log }_{10}}t} \over {1 + t}}dt}$ ...... (i)</p> <p>$$f\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt} $$</p> <p>Substituting $t \to {1 \over p}$</p> <p>$$f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}\left( {{1 \over p}} \right)} \over {1 + {1 \over p}}}\left( {{{ - 1} \over {{p^2}}}} \right)dp} $$</p> <p>$$ = \int_1^\alpha {{{{{\log }_{10}}p} \over {p(p + 1)}}dp = \int_1^\alpha {\left( {{{{{\log }_{10}}t} \over t} - {{{{\log }_{10}}t} \over {t + 1}}} \right)dt} } $$ ....... (ii)</p> <p>By (i) + (ii)</p> <p>$$f(\alpha ) + f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}t} \over t}dt = \int_1^\alpha {{{\ln t} \over t}\,.\,{{\log }_{10}}e\,dt} } $$</p> <p>$= {{{{(\ln \alpha )}^2}} \over {2{{\log }_e}10}}$</p> <p>$$\alpha = {e^3} \Rightarrow f({e^3}) + f({e^{ - 3}}) = {9 \over {2{{\log }_e}10}}$$</p>