The number of integers in the domain of function, satisfying `f(x)+f(x^(-1))=(x^3+1)/x ,i s_____`

Explanation

To solve the problem, we need to find the number of integers in the domain of the function defined by the equation:

\[ f(x) + f\left(\frac{1}{x}\right) = \frac{x^3 + 1}{x} \]

### Step-by-Step Solution:

1. **Rewrite the Equation:**
Start with the given equation:
\[ f(x) + f\left(\frac{1}{x}\right) = \frac{x^3 + 1}{x} \]
This can be simplified to:
\[ f(x) + f\left(\frac{1}{x}\right) = x^2 + \frac{1}{x} \]

2. **Substitute \( x \) with \( \frac{1}{x} \):**
Now, replace \( x \) with \( \frac{1}{x} \) in the original equation:
\[ f\left(\frac{1}{x}\right) + f(x) = \frac{\left(\frac{1}{x}\right)^3 + 1}{\frac{1}{x}} = \frac{\frac{1}{x^3} + 1}{\frac{1}{x}} = \frac{1 + x^3}{x^3} \]
This simplifies to:
\[ f\left(\frac{1}{x}\right) + f(x) = \frac{1}{x^2} + x \]

3. **Set Up the Equations:**
We now have two equations:
- Equation 1: \( f(x) + f\left(\frac{1}{x}\right) = x^2 + \frac{1}{x} \)
- Equation 2: \( f\left(\frac{1}{x}\right) + f(x) = \frac{1}{x^2} + x \)

Since the left-hand sides are the same, we can equate the right-hand sides:
\[ x^2 + \frac{1}{x} = \frac{1}{x^2} + x \]

4. **Rearranging the Equation:**
Rearranging gives us:
\[ x^2 - x + \frac{1}{x} - \frac{1}{x^2} = 0 \]
This can be rewritten as:
\[ x^2 - \frac{1}{x^2} + \frac{1}{x} - x = 0 \]

5. **Factoring the Equation:**
Recognize that \( x^2 - \frac{1}{x^2} \) can be factored using the difference of squares:
\[ (x - \frac{1}{x})(x + \frac{1}{x}) + \frac{1}{x} - x = 0 \]
Let \( y = x + \frac{1}{x} \). The equation becomes:
\[ (x - \frac{1}{x})(y) + \frac{1}{x} - x = 0 \]

6. **Finding Possible Values of \( x \):**
The equation \( x^2 = 1 \) gives us the possible solutions:
\[ x = 1 \quad \text{or} \quad x = -1 \]

7. **Determine the Domain:**
The integers in the domain of \( f(x) \) that satisfy the equation are \( x = 1 \) and \( x = -1 \).

### Final Answer:
Thus, the number of integers in the domain of the function is **2**.