Key Concepts and Formulas

• Definite Integral Properties: If $\int_a^b h(t) dt$ and $\int_a^b g(t) dt$ exist, then $\int_a^b h(t) dt + \int_a^b g(t) dt = \int_a^b (h(t) + g(t)) dt$.
• Substitution Rule for Definite Integrals: To evaluate $\int_a^b h(g(t)) g'(t) dt$, let $u = g(t)$, then $du = g'(t) dt$. The new limits of integration are $g(a)$ and $g(b)$.
• Logarithm Properties: $\ln(1/u) = -\ln u$.

Step-by-Step Solution

Step 1: Expressing $f(1/x)$ using a Substitution

We are given $f(x) = \int_1^x \frac{\ln t}{1+t} dt$. We need to find $f(1/x)$. $f\left(\frac{1}{x}\right) = \int_1^{1/x} \frac{\ln t}{1+t} dt$ To simplify this integral, we use the substitution $t = \frac{1}{u}$.

• Reasoning: This substitution is chosen because it often transforms integrals with limits involving $x$ and $1/x$ into integrals with limits involving $1$ and $x$, and it simplifies terms like $1+t$ and $\ln t$ in a useful way.
• Finding $dt$: Differentiating $t = u^{-1}$ with respect to $u$, we get $dt = -u^{-2} du = -\frac{1}{u^2} du$.
• Changing Limits:
  • When $t = 1$, $1 = \frac{1}{u} \implies u = 1$.
  • When $t = \frac{1}{x}$, $\frac{1}{x} = \frac{1}{u} \implies u = x$. Substituting these into the integral: $f\left(\frac{1}{x}\right) = \int_1^x \frac{\ln(1/u)}{1 + 1/u} \left(-\frac{1}{u^2}\right) du$ Now, we simplify the integrand:
• $\ln(1/u) = -\ln u$
• $1 + \frac{1}{u} = \frac{u+1}{u}$ Substituting these simplifications: $f\left(\frac{1}{x}\right) = \int_1^x \frac{-\ln u}{(u+1)/u} \left(-\frac{1}{u^2}\right) du$ $f\left(\frac{1}{x}\right) = \int_1^x \frac{-\ln u \cdot u}{u+1} \left(-\frac{1}{u^2}\right) du$ $f\left(\frac{1}{x}\right) = \int_1^x \frac{\ln u}{u(u+1)} du$ Since $u$ is a dummy variable, we can replace it with $t$: $f\left(\frac{1}{x}\right) = \int_1^x \frac{\ln t}{t(1+t)} dt$

Step 2: Combining $f(x)$ and $f(1/x)$

Now we add the expressions for $f(x)$ and $f(1/x)$. Since they have the same limits of integration, we can combine them under a single integral. $f(x) + f\left(\frac{1}{x}\right) = \int_1^x \frac{\ln t}{1+t} dt + \int_1^x \frac{\ln t}{t(1+t)} dt$ $f(x) + f\left(\frac{1}{x}\right) = \int_1^x \left(\frac{\ln t}{1+t} + \frac{\ln t}{t(1+t)}\right) dt$ Factor out $\ln t$: $f(x) + f\left(\frac{1}{x}\right) = \int_1^x \ln t \left(\frac{1}{1+t} + \frac{1}{t(1+t)}\right) dt$ Simplify the terms inside the parenthesis: $\frac{1}{1+t} + \frac{1}{t(1+t)} = \frac{t}{t(1+t)} + \frac{1}{t(1+t)} = \frac{t+1}{t(1+t)} = \frac{1}{t}$ Substitute this back into the integral: $f(x) + f\left(\frac{1}{x}\right) = \int_1^x \ln t \left(\frac{1}{t}\right) dt$ $f(x) + f\left(\frac{1}{x}\right) = \int_1^x \frac{\ln t}{t} dt$

Step 3: Evaluating the Simplified Integral

We need to evaluate the integral $\int_1^x \frac{\ln t}{t} dt$. We use the substitution $v = \ln t$.

• Reasoning: The derivative of $\ln t$ is $1/t$, which is present in the integrand, making this substitution ideal.
• Finding $dv$: Differentiating $v = \ln t$ with respect to $t$, we get $dv = \frac{1}{t} dt$.
• Changing Limits:
  • When $t = 1$, $v = \ln 1 = 0$.
  • When $t = x$, $v = \ln x$. Substitute these into the integral: $\int_1^x \frac{\ln t}{t} dt = \int_0^{\ln x} v \, dv$ Now, we evaluate this simple integral: $\int_0^{\ln x} v \, dv = \left[\frac{v^2}{2}\right]_0^{\ln x}$ $= \frac{(\ln x)^2}{2} - \frac{0^2}{2}$ $= \frac{1}{2}(\ln x)^2$ Thus, we have found that $f(x) + f(1/x) = \frac{1}{2}(\ln x)^2$.

Step 4: Calculating the Specific Value

The problem asks for the value of $f(e) + f(1/e)$. We use the general expression derived in Step 3 and substitute $x = e$. $f(e) + f\left(\frac{1}{e}\right) = \frac{1}{2}(\ln e)^2$ We know that $\ln e = 1$. $f(e) + f\left(\frac{1}{e}\right) = \frac{1}{2}(1)^2$ $f(e) + f\left(\frac{1}{e}\right) = \frac{1}{2}$

Common Mistakes & Tips

• Incorrect Substitution: Ensure the substitution correctly transforms both the differential ($dt$) and the limits of integration. A common error is forgetting to change the limits.
• Algebraic Errors: The simplification of $\frac{1}{1+t} + \frac{1}{t(1+t)}$ is crucial. Double-check algebraic manipulations to avoid errors.
• Logarithm Properties: Be careful with logarithm properties, especially $\ln(1/u) = -\ln u$.

Summary

The problem requires evaluating $f(e) + f(1/e)$ where $f(x)$ is defined by a definite integral. The strategy involves transforming the integral for $f(1/x)$ using the substitution $t=1/u$. This allows us to combine $f(x)$ and $f(1/x)$ into a single integral, which simplifies significantly after algebraic manipulation. The resulting integral $\int_1^x \frac{\ln t}{t} dt$ is then evaluated using a straightforward substitution. Finally, substituting $x=e$ into the general expression for $f(x) + f(1/x)$ yields the desired numerical value.

The final answer is \boxed{1/2}.