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 find the value of \( f(x) + f\left(\frac{1}{x}\right) \) where

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

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

We start by calculating \( f\left(\frac{1}{x}\right) \):

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

### Step 2: Change of Variables

We will use the substitution \( t = \frac{1}{u} \). Then, \( dt = -\frac{1}{u^2} \, du \).

When \( t = 1 \), \( u = 1 \) and when \( t = \frac{1}{x} \), \( u = x \).

Thus, we can rewrite the integral:

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

### Step 3: Simplify the Integral

Now simplify the integrand:

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

Thus, we have:

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

### Step 4: Combine the Integrals

Now we can combine \( f(x) \) and \( f\left(\frac{1}{x}\right) \):

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

### Step 5: Factor Out the Common Integral

Now we can factor out the common integral:

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

### Step 6: Evaluate the Integral

To evaluate the integral, we can use the property of logarithms and integrals:

\[ \int \frac{\log t}{1+t} \, dt = \frac{1}{2} \log^2 t + C. \]

Thus, we can evaluate:

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

### Step 7: Final Result

Putting it all together, we have:

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

Thus, the final answer is:

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