Let `f:(0,oo) in R` be given
`f(x)=overset(x)underset(1//x)int e^(t+(1)/(t))(1)/(t)dt`, then

Options

A. `f(x)` is monotonically increasing on`[1,oo)`
B. `f(x)` is monotonocally decreasing on`(0,1)`
C. `f(x)+f(1/x)=0`, for all `x epsilon (0,oo)`
D. `f(2^(x))` is an odd function of `x` on `R`

Explanation

To solve the given problem step by step, we will analyze the function defined by the integral and differentiate it to determine its properties.

### Step 1: Define the function
We start with the function defined as:
\[ f(x) = \int_{1/x}^{x} e^{t + \frac{1}{t}} \frac{1}{t} dt \]

### Step 2: Differentiate the function
To analyze the behavior of \( f(x) \), we will differentiate it with respect to \( x \) using the Leibniz rule for differentiation under the integral sign:
\[ f'(x) = \frac{d}{dx} \left( \int_{1/x}^{x} e^{t + \frac{1}{t}} \frac{1}{t} dt \right) \]
Using the Leibniz rule:
\[ f'(x) = e^{x + \frac{1}{x}} \cdot \frac{1}{x} - e^{1/x + x} \cdot \left(-\frac{1}{x^2}\right) \]
This simplifies to:
\[ f'(x) = e^{x + \frac{1}{x}} \cdot \frac{1}{x} + e^{1/x + x} \cdot \frac{1}{x^2} \]

### Step 3: Analyze the derivative
Both terms in \( f'(x) \) are positive for \( x > 0 \):
- \( e^{x + \frac{1}{x}} > 0 \)
- \( \frac{1}{x} > 0 \)
- \( e^{1/x + x} > 0 \)
- \( \frac{1}{x^2} > 0 \)

Thus, we conclude:
\[ f'(x) > 0 \quad \text{for } x > 0 \]
This indicates that \( f(x) \) is a monotonically increasing function on the interval \( (0, \infty) \).

### Step 4: Evaluate \( f(x) + f(1/x) \)
Next, we compute \( f(1/x) \):
\[ f\left(\frac{1}{x}\right) = \int_{x}^{1/x} e^{t + \frac{1}{t}} \frac{1}{t} dt \]
By changing the limits of integration, we have:
\[ f\left(\frac{1}{x}\right) = -\int_{1/x}^{x} e^{t + \frac{1}{t}} \frac{1}{t} dt = -f(x) \]
Thus:
\[ f(x) + f(1/x) = f(x) - f(x) = 0 \]

### Step 5: Conclusion
From the analysis, we have established:
1. \( f(x) \) is monotonically increasing on \( (0, \infty) \).
2. \( f(x) + f(1/x) = 0 \).

### Final Result
The properties of the function \( f(x) \) are:
- \( f(x) \) is monotonically increasing.
- \( f(x) + f(1/x) = 0 \).