If $$f(\alpha) = \int_1^\alpha \frac{\log_{10} t}{1+t} dt$$, $$\alpha > 0$$, then $$f(e^3) + f(e^{-3})$$ is equal to

We have $$f(\alpha) = \int_1^\alpha \frac{\log_{10} t}{1+t}\,dt$$ with $$\alpha > 0$$, and we need to compute $$f(e^3) + f(e^{-3})$$.

Since $$\log_{10} t = \frac{\ln t}{\ln 10}$$, we can write $$f(\alpha) = \frac{1}{\ln 10}\int_1^\alpha \frac{\ln t}{1+t}\,dt$$. Now in $$f(e^3)$$, we substitute $$t = e^s$$ so that $$dt = e^s\,ds$$, and when $$t$$ goes from 1 to $$e^3$$, $$s$$ goes from 0 to 3. This gives us $$f(e^3) = \frac{1}{\ln 10}\int_0^3 \frac{s\,e^s}{1+e^s}\,ds$$.

Similarly, in $$f(e^{-3})$$, we substitute $$t = e^{-s}$$ so that $$dt = -e^{-s}\,ds$$. When $$t$$ goes from 1 to $$e^{-3}$$, $$s$$ goes from 0 to 3. We get $$f(e^{-3}) = \frac{1}{\ln 10}\int_0^3 \frac{s\,e^{-s}}{1+e^{-s}}\cdot 1\,ds = \frac{1}{\ln 10}\int_0^3 \frac{s}{e^s+1}\,ds$$.

Now we add the two expressions:

$$f(e^3) + f(e^{-3}) = \frac{1}{\ln 10}\int_0^3 s\left(\frac{e^s}{1+e^s} + \frac{1}{1+e^s}\right)ds$$

We observe that $$\frac{e^s}{1+e^s} + \frac{1}{1+e^s} = \frac{e^s + 1}{1+e^s} = 1$$. This is the key simplification — the two fractions are complementary and sum to unity for every value of $$s$$.

Therefore $$f(e^3) + f(e^{-3}) = \frac{1}{\ln 10}\int_0^3 s\,ds = \frac{1}{\ln 10}\cdot\frac{s^2}{2}\Big|_0^3 = \frac{9}{2\ln 10} = \frac{9}{2\log_e 10}$$.

Hence, the correct answer is Option D.