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Question

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 :

9
$\frac{9}{2}$
$\frac{9}{\log _{e}(10)}$
$\frac{9}{2 \log _{e}(10)}$

Solution

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