`int_(0)^(infty)f(x+(1)/(x)) (ln x )/(x)dx` is equal to:

Options

A. (a) `0`
B. (b) `1`
C. (c) `(1)/(2)`
D. (d) cannot be evaluated

Answer

To solve the integral \( I = \int_{0}^{\infty} f\left(x + \frac{1}{x}\right) \frac{\ln x}{x} \, dx \), we will use a substitution method. Let's break it down step by step.

### Step 1: Define the integral
Let
\[ I = \int_{0}^{\infty} f\left(x + \frac{1}{x}\right) \frac{\ln x}{x} \, dx. \]

### Step 2: Make the substitution
We will use the substitution \( x = \frac{1}{u} \). Then, we have:
- \( dx = -\frac{1}{u^2} \, du \)
- When \( x \to 0 \), \( u \to \infty \)
- When \( x \to \infty \), \( u \to 0 \)

### Step 3: Change the limits and the integral
Substituting into the integral, we get:
\[ I = \int_{\infty}^{0} f\left(\frac{1}{u} + u\right) \frac{\ln\left(\frac{1}{u}\right)}{\frac{1}{u}} \left(-\frac{1}{u^2}\right) \, du. \]
This simplifies to:
\[ I = \int_{0}^{\infty} f\left(u + \frac{1}{u}\right) \frac{-\ln u}{u} \, du. \]

### Step 4: Simplify the logarithm
Using the property of logarithms, \( \ln\left(\frac{1}{u}\right) = -\ln u \), we can rewrite the integral as:
\[ I = \int_{0}^{\infty} f\left(u + \frac{1}{u}\right) \frac{\ln u}{u} \, du. \]

### Step 5: Recognize the symmetry
Notice that the integral \( I \) can be rewritten as:
\[ I = -\int_{0}^{\infty} f\left(u + \frac{1}{u}\right) \frac{\ln u}{u} \, du. \]
Thus, we have:
\[ I = -I. \]

### Step 6: Solve for \( I \)
Adding \( I \) to both sides gives:
\[ 2I = 0 \implies I = 0. \]

### Conclusion
The value of the integral is:
\[ I = 0. \]

### Final Answer
Thus, the integral evaluates to \( \boxed{0} \).

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