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