1. We are given

$f(\alpha)=\int_1^{\alpha} \frac{\log_{10} t}{1+t}\,dt, \qquad \alpha>0$

and we need to find

$f(e^3)+f(e^{-3}).$

2. First, write $\log_{10} t$ in terms of natural logarithm:

$\log_{10} t=\frac{\ln t}{\ln 10}.$

So,

$f(\alpha)=\frac{1}{\ln 10}\int_1^{\alpha}\frac{\ln t}{1+t}\,dt.$

Hence,

$f(e^3)+f(e^{-3})=\frac{1}{\ln 10}\left(\int_1^{e^3}\frac{\ln t}{1+t}\,dt+\int_1^{e^{-3}}\frac{\ln t}{1+t}\,dt\right).$

3. Let

$I=\int_1^{e^3}\frac{\ln t}{1+t}\,dt+\int_1^{e^{-3}}\frac{\ln t}{1+t}\,dt.$

Now rewrite the second integral by reversing limits:

$\int_1^{e^{-3}}\frac{\ln t}{1+t}\,dt=-\int_{e^{-3}}^1\frac{\ln t}{1+t}\,dt.$

Thus,

$I=\int_1^{e^3}\frac{\ln t}{1+t}\,dt-\int_{e^{-3}}^1\frac{\ln t}{1+t}\,dt.$

4. In the second integral, use the substitution

$t=\frac{1}{x} \quad \Rightarrow \quad dt=-\frac{1}{x^2}dx, \qquad \ln t=-\ln x.$

Also,

$1+t=1+\frac{1}{x}=\frac{x+1}{x}.$

Therefore,

$\frac{\ln t}{1+t}dt=\frac{-\ln x}{(x+1)/x}\left(-\frac{1}{x^2}dx\right)=\frac{\ln x}{x(x+1)}dx.$

The limits change as:

• when $t=e^{-3}$, $x=e^3$
• when $t=1$, $x=1$

So,

$\int_{e^{-3}}^1\frac{\ln t}{1+t}\,dt=\int_{e^3}^1 \frac{\ln x}{x(x+1)}dx=-\int_1^{e^3}\frac{\ln x}{x(x+1)}dx.$

Hence,

$I=\int_1^{e^3}\frac{\ln t}{1+t}\,dt+\int_1^{e^3}\frac{\ln t}{t(1+t)}\,dt.$

5. Combine the integrands:

$\frac{\ln t}{1+t}+\frac{\ln t}{t(1+t)}=\ln t\left(\frac{1}{1+t}+\frac{1}{t(1+t)}\right).$

Now,

$\frac{1}{1+t}+\frac{1}{t(1+t)}=\frac{t+1}{t(1+t)}=\frac{1}{t}.$

So,

$I=\int_1^{e^3}\frac{\ln t}{t}\,dt.$

6. Evaluate this integral:

$\int \frac{\ln t}{t}\,dt=\frac{(\ln t)^2}{2}.$

Therefore,

$I=\left[\frac{(\ln t)^2}{2}\right]_1^{e^3}=\frac{(\ln e^3)^2}{2}-\frac{(\ln 1)^2}{2}.$

Since $\ln e^3=3$ and $\ln 1=0$,

$I=\frac{3^2}{2}=\frac{9}{2}.$

7. Now multiply by $\frac{1}{\ln 10}$:

$f(e^3)+f(e^{-3})=\frac{1}{\ln 10}\cdot \frac{9}{2}=\frac{9}{2\ln 10}.$

Since $\ln 10=\log_e 10$,

$f(e^3)+f(e^{-3})=\frac{9}{2\log_e 10}.$

8. Comparing with the options, this is option D.