Integral and Derivative by J. S

$\begin{aligned} f(x) & = \left(\int_0^x e^{-t^2} dt \right)^2 \\ \implies f'(x) & = 2 \int_0^x e^{-t^2} dt \cdot e^{-x^2} \\ & = 2e^{-x^2} \int_0^x e^{-t^2} dt \end{aligned}$

$\begin{aligned} g(x) & = \int_0^1 \frac {e^{-x^2(t^2+1)}}{t^2+1} dt \\ \implies g'(x) & = \int_0^1 \frac {-2x(t^2+1)e^{-x^2(t^2+1)}}{t^2+1} dt \\ & = -2xe^{-x^2} \int_0^1 e^{-x^2t^2} dt & \small \color{#3D99F6} \text{Let }u = xt \implies du = x dt \\ & = -2xe^{-x^2} \int_0^x e^{-u^2} du & \small \color{#3D99F6} \text{Replace }u \text{ with }t \\ & = -2xe^{-x^2} \int_0^x e^{-t^2} dt \end{aligned}$

Note that $f'(x) + g'(x) = \displaystyle 2e^{-x^2} \int_0^x e^{-t^2} dt - 2e^{-x^2} \int_0^x e^{-t^2} dt = \boxed{0}$ for all $x$ .