Answer
$$ - \frac{1}{{{x^2} + 1}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\int_x^0 {\frac{{dp}}{{{p^2} + 1}}} \cr
& {\text{Property }}\int_a^b {f\left( u \right)} du = - \int_b^a {f\left( u \right)} du \cr
& = - \frac{d}{{dx}}\int_0^x {\frac{{dp}}{{{p^2} + 1}}} \cr
& {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr
& A\left( x \right) = \int_a^x {f\left( p \right)dp{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( p \right)dp} = f\left( x \right) \cr
& {\text{In this exercise}} \cr
& f\left( p \right) = \frac{1}{{{p^2} + 1}},{\text{ and }}a = 0 \cr
& then \cr
& - \frac{d}{{dx}}\int_0^x {\frac{{dp}}{{{p^2} + 1}}} = - \frac{1}{{{{\left( x \right)}^2} + 1}} \cr
& = - \frac{1}{{{x^2} + 1}} \cr} $$
