The domain of the function `f(x) = (x^(2) + 1)/(ln (x^(2) + 1))` is

Options

A. `[0, oo)`
B. `(-oo, 0]`
C. `(-oo, oo)`
D. `(-oo, oo) - {0}`

Answer

To find the domain of the function \( f(x) = \frac{x^2 + 1}{\ln(x^2 + 1)} \), we need to ensure that the denominator does not equal zero and that the argument of the logarithm is positive.

### Step-by-Step Solution:

1. **Identify the function**:
We have \( f(x) = \frac{x^2 + 1}{\ln(x^2 + 1)} \).

2. **Check the denominator**:
The denominator is \( \ln(x^2 + 1) \). For the function to be defined, this must not equal zero:
\[ \ln(x^2 + 1) \neq 0 \]

3. **Set the logarithm equal to zero**:
The logarithm equals zero when its argument equals 1:
\[ x^2 + 1 = 1 \]
Solving this gives:
\[ x^2 = 0 \quad \Rightarrow \quad x = 0 \]
Thus, \( x \) cannot be 0.

4. **Check the argument of the logarithm**:
The argument of the logarithm must be positive:
\[ x^2 + 1 > 0 \]
Since \( x^2 \) is always non-negative (it is zero or positive), adding 1 ensures that:
\[ x^2 + 1 \geq 1 > 0 \]
This condition is satisfied for all real numbers \( x \).

5. **Combine the conditions**:
The only restriction we have is that \( x \) cannot be 0. Therefore, the domain of \( f(x) \) is all real numbers except 0.

### Conclusion:
The domain of the function \( f(x) = \frac{x^2 + 1}{\ln(x^2 + 1)} \) is:
\[ \text{Domain} = \mathbb{R} \setminus \{0\} \]