Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 406: 9

$$g'(x) = \sqrt{x + x^3}$$

$$g(x) = \int\limits_0^x{\sqrt{t + t^3}}dt$$ Plug in the upper bound into t and take the derivative of the upper bound. $$g'(x) = \sqrt{(x) + (x)^3} \times (x)'$$ Simplify $$g'(x) = \sqrt{x + x^3}$$