To find the domain of the function $f(x) = \ln(x^2 + 1)$, we need to determine the values of $x$ for which the argument of the logarithm, $x^2 + 1$, is positive. The logarithm function is defined only when its argument is greater than 0:

$x^2 + 1 > 0$

Since $x^2 \geq 0$ for all real $x$, adding 1 ensures that $x^2 + 1$ is always positive. Therefore, there are no restrictions on $x$ based on this inequality.

Domain:

The domain of $f(x) = \ln(x^2 + 1)$ is all real numbers:

$\text{Domain: } (-\infty, \infty)$

Would you like more details or have any other questions?

Here are some related questions to further explore the topic:

1. How would the domain change if the function was $\ln(x^2 - 1)$?
2. What is the range of the function $f(x) = \ln(x^2 + 1)$?
3. How does the derivative of $f(x) = \ln(x^2 + 1)$ look?
4. What is the behavior of $f(x) = \ln(x^2 + 1)$ as $x$ approaches infinity?
5. How does the function behave for negative values of $x$?

Tip: When determining the domain of logarithmic functions, always ensure the argument of the logarithm is strictly positive.