Is there any special relation between the integrals
$$f(x)=\left(\int_{0}^{x}e^{-t^2} dt\right)^2,g(x)=\int_{0}^{1}\frac{e^{-x^2(t^2+1)}}{t^2+1} dt$$
as these two integrals can be used to show that, $\int_{0}^{\infty}e^{-t^2} dt=\frac{\sqrt{\pi}}{2}$
Your question is about the third proof here. Since$$f^\prime(x)=-2e^{-x^2}\int_0^xe^{-t^2}dt\stackrel{y:=\frac{t}{x}}{=}-\int_0^12xe^{-(1+y^2)x^2}dy=-\frac{d}{dx}\int_0^1\frac{e^{(1+y^2)x^2}}{1+y^2}dy=-g^\prime(x),$$$f+g$ is constant. Its value with $x=\infty$ is the square of the integral you want to evaluate; its value with $x=0$ is $\int_0^1\frac{dt}{t^2+1}=\frac{\pi}{4}$.