To solve the problem using Part 1 of the Fundamental Theorem of Calculus, we proceed as follows:

Step 1: Recall Part 1 of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus, Part 1, states that if $F(x)$ is defined as: $F(x) = \int_a^x f(t) \, dt,$ then the derivative of $F(x)$ with respect to $x$ is: $F'(x) = f(x),$ provided $f(t)$ is continuous on the interval.

Step 2: Apply the Theorem to the Given Function

The function $g(x)$ is defined as: $g(x) = \int_1^x \ln(2 + t^2) \, dt.$

Here:

• The lower limit of integration is constant ($1$).
• The upper limit of integration is $x$, which is a variable.

By Part 1 of the Fundamental Theorem: $g'(x) = \ln(2 + x^2).$

Final Answer:

$g'(x) = \ln(2 + x^2).$

This result follows because the derivative simply evaluates the integrand at the upper limit of integration.

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Would you like a deeper explanation, or do you have any questions? Here are some related questions to expand your understanding:

1. How does the Fundamental Theorem of Calculus relate to antiderivatives?
2. What happens if the upper limit is a more complex function (e.g., $x^2$)?
3. Can this theorem apply to integrals with variable lower limits instead?
4. What are the continuity requirements for using this theorem?
5. How do we handle problems where both limits of integration depend on $x$?

Tip: Remember, the derivative of a definite integral where the variable appears as the upper limit is simply the integrand evaluated at that limit!