Key Concepts and Formulas

1. Properties of Definite Integrals: The value of a definite integral is independent of the variable of integration ($\int_a^b f(x) dx = \int_a^b f(t) dt$). Integrals with the same limits can be added or subtracted by combining their integrands.
2. Substitution Rule for Definite Integrals: When performing a substitution $u = g(x)$, the limits of integration must be transformed accordingly. For $\int_a^b f(x) dx$, if $u=g(x)$, the new limits are $g(a)$ and $g(b)$.
3. Fundamental Theorem of Calculus: $\int_a^b h'(x) dx = h(b) - h(a)$.
4. Logarithm Properties: $\log(1/x) = -\log x$.

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Step-by-Step Solution

We are given the function $F(x) = f(x) + f(1/x)$, where $f(x) = \int_1^x \frac{\log t}{1+t} dt$. We need to find the value of $F(e)$.

Step 1: Express $F(e)$ in terms of $f(e)$ and $f(1/e)$. Substitute $x=e$ into the definition of $F(x)$: $F(e) = f(e) + f\left(\frac{1}{e}\right)$ Now, use the definition of $f(x)$ to write $f(e)$ and $f(1/e)$ as definite integrals: $f(e) = \int_1^e \frac{\log t}{1+t} dt$ $f\left(\frac{1}{e}\right) = \int_1^{1/e} \frac{\log t}{1+t} dt$ Therefore, $F(e) = \int_1^e \frac{\log t}{1+t} dt + \int_1^{1/e} \frac{\log t}{1+t} dt \quad \ldots(1)$

Step 2: Evaluate the second integral using a substitution. Let the second integral be $I = \int_1^{1/e} \frac{\log t}{1+t} dt$. We will use the substitution $z = 1/t$. From $z = 1/t$, we have $t = 1/z$. Differentiating with respect to $z$, we get $dt = -\frac{1}{z^2} dz$. Now, we change the limits of integration: When $t = 1$, $z = 1/1 = 1$. When $t = 1/e$, $z = 1/(1/e) = e$.

Substitute these into the integral $I$: $I = \int_1^e \frac{\log(1/z)}{1 + 1/z} \left(-\frac{1}{z^2}\right) dz$ Using the logarithm property $\log(1/z) = -\log z$: $I = \int_1^e \frac{-\log z}{\frac{z+1}{z}} \left(-\frac{1}{z^2}\right) dz$ $I = \int_1^e \frac{-\log z \cdot z}{z+1} \left(-\frac{1}{z^2}\right) dz$ $I = \int_1^e \frac{\log z}{z(z+1)} dz$ Since the variable of integration does not affect the value of a definite integral, we can replace $z$ with $t$: $I = \int_1^e \frac{\log t}{t(t+1)} dt$

Step 3: Combine the integrals for $F(e)$. Substitute the evaluated form of $I$ back into equation (1): $F(e) = \int_1^e \frac{\log t}{1+t} dt + \int_1^e \frac{\log t}{t(1+t)} dt$ Since the limits of integration are the same, we can combine the integrands: $F(e) = \int_1^e \left(\frac{\log t}{1+t} + \frac{\log t}{t(1+t)}\right) dt$ Find a common denominator for the terms inside the parenthesis: $F(e) = \int_1^e \left(\frac{t \log t}{t(1+t)} + \frac{\log t}{t(1+t)}\right) dt$ $F(e) = \int_1^e \frac{t \log t + \log t}{t(1+t)} dt$ Factor out $\log t$ from the numerator: $F(e) = \int_1^e \frac{\log t (t+1)}{t(1+t)} dt$ For $t \in [1, e]$, $t+1 \neq 0$, so we can cancel the $(t+1)$ term: $F(e) = \int_1^e \frac{\log t}{t} dt$

Step 4: Evaluate the final integral. To evaluate $\int_1^e \frac{\log t}{t} dt$, we use the substitution $u = \log t$. Differentiating with respect to $t$, we get $du = \frac{1}{t} dt$. Now, we change the limits of integration: When $t = 1$, $u = \log 1 = 0$. When $t = e$, $u = \log e = 1$.

Substitute $u$ and $du$ into the integral: $F(e) = \int_0^1 u \, du$ This is a standard integral: $F(e) = \left[\frac{u^2}{2}\right]_0^1$ Apply the Fundamental Theorem of Calculus: $F(e) = \frac{1^2}{2} - \frac{0^2}{2}$ $F(e) = \frac{1}{2} - 0$ $F(e) = \frac{1}{2}$

Common Mistakes & Tips

• Limit Transformation: Always remember to change the limits of integration when performing a substitution in a definite integral.
• Algebraic Simplification: Be meticulous with algebraic manipulations, especially when combining fractions and canceling terms. Errors here can lead to a completely wrong result.
• Recognizing Patterns: The structure of $F(x) = f(x) + f(1/x)$ is a strong indicator that a substitution of the form $z=1/t$ will be beneficial.

Summary

The problem involves evaluating $F(e) = f(e) + f(1/e)$, where $f(x)$ is defined by an integral. By expressing $F(e)$ as a sum of two integrals and applying the substitution $z=1/t$ to the integral involving $f(1/e)$, we transformed the problem into a sum of integrals with the same limits. Algebraic simplification led to a single integral $\int_1^e \frac{\log t}{t} dt$. This integral was then easily evaluated using the substitution $u = \log t$, yielding the result $1/2$.

The final answer is $\boxed{1/2}$, which corresponds to option (C).