Let `F(x) =f(x) +f((1)/(x)),"where" f(x)=int_(1)^(x) (log t)/(1+t) dt` Then F (e) equals

Options

A. 1
B. 2
C. `1//2`
D. 0

Explanation

To solve the problem, we need to find \( F(e) \) where \( F(x) = f(x) + f\left(\frac{1}{x}\right) \) and \( f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt \).

### Step 1: Find \( f\left(\frac{1}{x}\right) \)

We start by calculating \( f\left(\frac{1}{x}\right) \):
\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{\log t}{1+t} dt \]

To evaluate this integral, we will use a substitution. Let \( t = \frac{1}{u} \), then \( dt = -\frac{1}{u^2} du \). The limits change as follows:
- When \( t = 1 \), \( u = 1 \)
- When \( t = \frac{1}{x} \), \( u = x \)

Thus, we have:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log\left(\frac{1}{u}\right)}{1+\frac{1}{u}} \left(-\frac{1}{u^2}\right) du \]

### Step 2: Simplify the Integral

Now, we simplify the integrand:
\[ \log\left(\frac{1}{u}\right) = -\log u \]
\[ 1 + \frac{1}{u} = \frac{u+1}{u} \]
Thus, the integrand becomes:
\[ \frac{-\log u}{\frac{u+1}{u}} \left(-\frac{1}{u^2}\right) = \frac{\log u}{1 + u} du \]

So we have:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log u}{1+u} du \]

### Step 3: Combine \( f(x) \) and \( f\left(\frac{1}{x}\right) \)

Now, we can express \( F(x) \):
\[ F(x) = f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log t}{1+t} dt + \int_{1}^{x} \frac{\log u}{1+u} du \]

Since both integrals have the same limits and integrands, we can combine them:
\[ F(x) = \int_{1}^{x} \frac{\log t}{1+t} dt + \int_{1}^{x} \frac{\log t}{1+t} dt = 2 \int_{1}^{x} \frac{\log t}{1+t} dt \]

### Step 4: Evaluate \( F(e) \)

Now we need to evaluate \( F(e) \):
\[ F(e) = 2 \int_{1}^{e} \frac{\log t}{1+t} dt \]

### Step 5: Calculate \( f(x) \)

We can find \( f(x) \) directly:
\[ f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt \]

Using integration by parts or known results, we find:
\[ f(x) = \frac{1}{2} \log^2 x \]

### Step 6: Substitute \( x = e \)

Now substituting \( x = e \):
\[ f(e) = \frac{1}{2} \log^2 e = \frac{1}{2} (1)^2 = \frac{1}{2} \]

Thus:
\[ F(e) = f(e) + f\left(\frac{1}{e}\right) = f(e) + f(e) = 2f(e) = 2 \cdot \frac{1}{2} = 1 \]

### Final Answer
\[ F(e) = 1 \]