If $\mathrm{f}\left(\mathrm{\alpha }\right)={\int }_1^{\mathrm{\alpha }} \frac{{\log }_{10}\mathrm{t}}{1+\mathrm{t}}\mathrm{dt},\mathrm{\alpha }>0,\text{ then }\mathrm{f}\left({\mathrm{e}}^3\right)+\mathrm{f}\left({\mathrm{e}}^{-3}\right)$ is equal to 

$\mathrm{f}\left(\mathrm{\alpha }\right)={\int }_1^{\mathrm{\alpha }} \frac{{\log }_{10}\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}\dots \left(\mathrm{i}\right);\mathrm{f}\left({\mathrm{e}}^3\right)={\int }_1^{{\mathrm{e}}^3} \frac{{\log }_{10}\mathrm{t}}{1+\mathrm{t}}\mathrm{dt} ....\left(\mathrm{ii}\right)$
Put $\mathrm{t}=\frac{1}{\mathrm{x}}\text{ in }\left(\mathrm{i}\right)$, we get
$\mathrm{f}\left(\mathrm{\alpha }\right)={\int }_1^{1/\mathrm{\alpha }} \frac{{\log }_{10}(1/\mathrm{x})}{1+(1/\mathrm{x})}\left(\frac{-1}{{\mathrm{x}}^2}\right)\mathrm{dx}={\int }_1^{1/\mathrm{\alpha }} \frac{{\log }_{10}\mathrm{x}}{\mathrm{x}(\mathrm{x}+1)}\mathrm{dx}$
$\therefore \mathrm{f}\left({\mathrm{e}}^{-3}\right)={\int }_1^{{\mathrm{e}}^3} \frac{{\log }_{10}\mathrm{x}}{\mathrm{x}(\mathrm{x}+1)}\mathrm{dx}={\int }_1^{{\mathrm{e}}^3} \frac{{\log }_{10}\mathrm{t}}{\mathrm{t}(\mathrm{t}+1)}\mathrm{dt} ....\left(\mathrm{iii}\right)$
Adding (ii) and (iii) , we get 
$\mathrm{f}\left({\mathrm{e}}^3\right)+\mathrm{f}\left({\mathrm{e}}^{-3}\right)={\int }_1^{{\mathrm{e}}^3} \frac{{\log }_{10}\mathrm{t}}{\mathrm{t}+1}\left(1+\frac{1}{\mathrm{t}}\right)\mathrm{dt}={\int }_1^{{\mathrm{e}}^3} \frac{{\log }_{10}\mathrm{t}}{\mathrm{t}}\mathrm{dt}$
Put ${\log }_{10}\mathrm{t}=\mathrm{z}\Rightarrow \frac{1}{{\log }_{\mathrm{e}}10}\cdot \frac{1}{\mathrm{t}}\mathrm{dt}=\mathrm{dz}$
$\mathrm{t}=1\Rightarrow \mathrm{z}=0,\mathrm{t}={\mathrm{e}}^3\Rightarrow \mathrm{z}=3/{\log }_{\mathrm{e}}10$
$\begin{array}{l}\therefore \mathrm{f}\left({\mathrm{e}}^3\right)+\mathrm{f}\left({\mathrm{e}}^{-3}\right)={\int }_0^{3/{\log }_{\mathrm{e}}10} \mathrm{z}\cdot {\log }_{\mathrm{e}}10\mathrm{dz}\\ ={\log }_{\mathrm{e}}10\cdot {\left[\frac{{\mathrm{z}}^2}{2}\right]}_0^{3/{\log }_{\mathrm{e}}10}=\frac{{\log }_{\mathrm{e}}10}{2}\left(\frac{9}{{\left({\log }_{\mathrm{e}}10\right)}^2}-0\right)\\ =\frac{9}{2{\log }_{\mathrm{e}}10}\end{array}$