Answer
$${x^2} + x + 1$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\int_3^x {\left( {{t^2} + t + 1} \right)dt} \cr
& {\text{Using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr
& A\left( x \right) = \int_a^x {f\left( t \right)dt{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( t \right)dt} = f\left( x \right) \cr
& {\text{In this exercise}} \cr
& f\left( t \right) = {t^2} + t + 1,{\text{ and }}a = 3 \cr
& then \cr
& \frac{d}{{dx}}\int_3^x {\left( {{t^2} + t + 1} \right)dt} = {\left( x \right)^2} + \left( x \right) + 1 \cr
& \frac{d}{{dx}}\int_3^x {\left( {{t^2} + t + 1} \right)dt} = {x^2} + x + 1 \cr} $$
