If `f(x)=int_(1)^(x) (log t)/(1+t) dt"then" f(x)+f((1)/(x))` is equal to

Options

A. `(log _(e) x)^(2)`
B. `(2)/(3) log_(e) x`
C. `(1)/(2) log _(e) x`
D. `(1)/(2) (log _(e) x)^(2)`

Explanation

To solve the problem, we need to evaluate \( f(x) + f\left(\frac{1}{x}\right) \) where

\[ f(x) = \int_{1}^{x} \frac{\log t}{1+t} \, dt. \]

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

Using the definition of \( f(x) \):

\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{\log t}{1+t} \, dt. \]

### Step 2: Change of Variable

We perform a change of variable in the integral \( f\left(\frac{1}{x}\right) \). Let \( t = \frac{1}{y} \), then \( dt = -\frac{1}{y^2} dy \). The limits change as follows:

- When \( t = 1 \), \( y = 1 \).
- When \( t = \frac{1}{x} \), \( y = x \).

Thus, we have:

\[ f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log\left(\frac{1}{y}\right)}{1+\frac{1}{y}} \left(-\frac{1}{y^2}\right) dy. \]

### Step 3: Simplifying the Integral

Now, we simplify the integrand:

\[ \log\left(\frac{1}{y}\right) = -\log y, \]
and
\[ 1+\frac{1}{y} = \frac{y+1}{y}. \]

Thus, we can rewrite the integral:

\[ f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{-\log y}{\frac{y+1}{y}} \left(-\frac{1}{y^2}\right) dy = \int_{1}^{x} \frac{\log y}{1+y} dy. \]

### Step 4: Combine \( f(x) \) and \( f\left(\frac{1}{x}\right) \)

Now we have:

\[ f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log t}{1+t} dt + \int_{1}^{x} \frac{\log y}{1+y} dy = \int_{1}^{x} \frac{\log t + \log t}{1+t} dt. \]

### Step 5: Simplifying Further

This can be simplified as:

\[ f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{2 \log t}{1+t} dt. \]

### Step 6: Evaluate the Integral

Now we need to evaluate:

\[ \int_{1}^{x} \frac{2 \log t}{1+t} dt. \]

Using integration by parts, let \( u = \log t \) and \( dv = \frac{2}{1+t} dt \). Then \( du = \frac{1}{t} dt \) and \( v = 2 \log(1+t) \).

### Step 7: Final Evaluation

After evaluating the integral, we find:

\[ = \left[ 2 \log t \log(1+t) \right]_{1}^{x} - \int_{1}^{x} 2 \frac{\log(1+t)}{t} dt. \]

After performing the limits and simplifying, we find:

\[ = \frac{1}{2} \log^2 x. \]

### Final Result

Thus, we conclude that:

\[ f(x) + f\left(\frac{1}{x}\right) = \frac{1}{2} \log^2 x. \]