Animated Solution for Mathematics - Definite Integration: 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 are tasked with evaluating the expression $f(e^3)+f(e^{-3})$ for the function defined as:

At first glance, attempting to integrate this directly is a trap. The key to solving this problem lies in recognizing the relationship between $f(\alpha )$ and $f(1/\alpha )$.

First, we simplify the integrand by converting the base of the logarithm to the natural logarithm using the change of base formula, ${\log }_{10}t=\frac{\ln t}{\ln 10}$. Since $\ln 10$ is a constant, we pull it out:

We now consider the sum $f(\alpha )+f(1/\alpha )$. The second term involves an integral from $1$ to $1/\alpha$. To align the limits, we perform the substitution $t=1/x$.

For the integral $I_2={\int }_1^{1/\alpha }\frac{\ln t}{1+t}dt$, we set $t=1/x$, which implies $dt=-1/x^{2}dx$. The limits change from $t=1\to x=1$ and $t=1/\alpha \to x=\alpha$.

Substituting these into the integral yields:

Simplifying the terms inside the integral, we note that $\ln (1/x)=-\ln x$ and $1+1/x=(x+1)/x$. The two negative signs cancel, and the expression simplifies beautifully:

We now combine the two integrals under the common factor $\frac{1}{\ln 10}$:

Factoring out $\ln x$ and simplifying the sum of the fractions:

The expression reduces to a simple integral:

Evaluating this integral using the substitution $u=\ln x$, we obtain $\frac{1}{2}(\ln \alpha {)}^2$. Substituting $\alpha =e^3$ and $\ln (e^3)=3$:

The final result is $\frac{9}{2\ln 10}$.