\( f(\alpha)=\displaystyle\int\limits_{1}^{\alpha}\frac{\log_{10} t}{1+t}\,dt, \ \alpha>0 \) If , then \( f(e^{3})+f(e^{-3}) \) is equal to:
\( 9 \)
\( \dfrac{9}{2} \)
\( \dfrac{9}{\log_{e}(10)} \)
\( \dfrac{9}{2\log_{e}(10)} \)