## Calculus: Early Transcendentals 8th Edition

$$g'(x) = ln(1 + x^2)$$
$$g(x) = \int\limits_1^x {ln(1+t^2)}dt$$ 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. $$g'(x) = ln(1 + x^2)$$