The expression $e^{\ln(x^2 + 1)}$ simplifies as follows:

By the property of logarithms and exponentials, $e^{\ln(a)} = a$ for any $a > 0$. So, applying this to your expression:

$e^{\ln(x^2 + 1)} = x^2 + 1$

Thus, the simplified form of $e^{\ln(x^2 + 1)}$ is:

$x^2 + 1$

Would you like a more detailed explanation, or is there something specific you'd like to clarify?

Here are some related questions:

1. What other logarithmic and exponential properties are useful in simplifying expressions?
2. How would you approach the derivative of $x^2 + 1$?
3. Can this property be used with other logarithms, like $\ln(x+1)$?
4. What happens if you have $e^{\ln(f(x))}$ for a more complex function $f(x)$?
5. How would you apply this simplification in solving integrals?

Tip: When you encounter an expression like $e^{\ln(x)}$, remember it simplifies directly to $x$. This can help save time when solving problems!