1. The problem is to simplify or understand the expression $\sqrt{x^2 + 1}$.\n\n2. Inside the square root, we have $x^2 + 1$. Since $x^2$ is always non-negative and 1 is positive, the expression inside the root is always positive, so the square root is defined for all real $x$.\n\n3. The expression $\sqrt{x^2 + 1}$ cannot be simplified further into elementary functions because $x^2 + 1$ is not a perfect square.\n\n4. This function represents the distance from the origin to the point $(x,1)$ on the Cartesian plane, or the hypotenuse of a right triangle with legs $x$ and 1.\n\n5. The function is always greater than or equal to 1, since $x^2 \geq 0$ implies $x^2 + 1 \geq 1$.\n\nFinal answer: The expression $\sqrt{x^2 + 1}$ is already in simplest form and is defined for all real $x$.