The function f(x) = log (x² + √x2+1) is

The correct option is "even function".

Let us analyze the given function step-by-step to understand why it is classified as an even function.
The given function is:

$$f(x)=\log (x^2+\sqrt{x^2+1})$$

To determine whether a function is even, odd, or neither, we evaluate the function at $-x$ and compare it with the original function:
1. A function is even if $f(-x)=f(x)$ for all $x$ in its domain.
2. A function is odd if $f(-x)=-f(x)$ for all $x$ in its domain.

Let us substitute $-x$ in place of $x$ in the expression for $f(x)$:

$$f(-x)=\log ({(-x)}^2+\sqrt{{(-x)}^2+1})$$

Since the square of a negative number is positive, we know that ${(-x)}^2=x^2$. Simplifying the terms inside the logarithm, we get:

$$f(-x)=\log (x^2+\sqrt{x^2+1})$$

Notice that this is exactly the original function definition:

$$f(-x)=f(x)$$

Since the condition $f(-x)=f(x)$ is satisfied, the function is indeed an even function.