MathematicsMedium210×since 2002Q3446If f(α)=∫1αlog10t1+tdt,α>0f(\alpha)=\int\limits_{1}^{\alpha} \frac{\log _{10} \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}, \alpha>0f(α)=1∫α1+tlog10tdt,α>0, then f(e3)+f(e−3)f\left(\mathrm{e}^{3}\right)+f\left(\mathrm{e}^{-3}\right)f(e3)+f(e−3) is equal to :A9B92\frac{9}{2}29C9loge(10)\frac{9}{\log _{e}(10)}loge(10)9D92loge(10)\frac{9}{2 \log _{e}(10)}2loge(10)9Check answerSkip