Question

use part one of the fundamental theorem of calculus to find the derivative of the function.
$g(x) = \int_{1}^{x} \ln(2 + t^2) \\, dt$
$g(x) = \ln\left(2 + x^2\
ight)2x$

Explanation:

Step1: Recall Fundamental Theorem of Calculus Part 1

The Fundamental Theorem of Calculus Part 1 states that if \( g(x)=\int_{a}^{x}f(t)dt \), then \( g'(x) = f(x) \), where \( a \) is a constant and \( f \) is continuous on the interval containing \( a \) and \( x \).

Step2: Apply the Theorem to the Given Function

For the function \( g(x)=\int_{1}^{x}\ln(2 + t^{2})dt \), we identify \( f(t)=\ln(2 + t^{2}) \). By the Fundamental Theorem of Calculus Part 1, we substitute \( t = x \) into \( f(t) \) to find \( g'(x) \). So \( g'(x)=\ln(2 + x^{2}) \).

Answer:

\( \ln(2 + x^{2}) \)