Answer
$$g'(x) = ln(1 + x^2)$$
Work Step by Step
$$g(x) = \int\limits_1^x {ln(1+t^2)}dt$$
Plug the upper bound into t and due to the chain rule,multiply it by the derivative of the upper bound.
$$g'(x) = ln(1 + (x)^2) \times (x)'$$
Simplify.
$$g'(x) = ln(1 + x^2)$$