Stuck on crucial step while computing $\int_{- \infty}^{\infty} e^{-t^2}dt$

$$F(x) = \int_0^1 dt \frac{e^{-x^2 (1+t^2)}}{1+t^2} \implies F'(x) = -2 x e^{-x^2} \int_0^1 dt \, e^{-x^2 t^2} $$

$$G(x) = \left ( \int_0^x dt \, e^{-t^2} \right )^2 = x^2 \left ( \int_0^1 dt \, e^{-x^2 t^2} \right )^2 \\ \implies G'(x) = 2 x \left ( \int_0^1 dt \, e^{-x^2 t^2} \right ) \left [\int_0^1 dt \, e^{-x^2 t^2} - 2 x^2 \int_0^1 dt \,t^2 e^{-x^2 t^2} \right ]$$

Then

$$F'(x) + G'(x) = 2 x \left ( \int_0^1 dt \, e^{-x^2 t^2} \right )\left [-e^{-x^2} + \int_0^1 dt \,(1-2 x^2 t^2) e^{-x^2 t^2} \right ] $$

Then note that

$$(1-2 x^2 t^2) e^{-x^2 t^2} = \frac{d}{dt} \left ( t e^{-x^2 t^2} \right )$$

and you'll find that $F'(x) + G'(x) = 0$ identically. Thus $F(x) = G(x)$ equals a constant, say $F(0)+G(0)$, which is

$$\int_0^1 \frac{dt}{1+t^2} = \frac{\pi}{4} $$

Now take the limit as $x \to \infty$, in which limit $F(x) \to 0$ and thus

$$\lim_{x \to \infty} \left ( \int_0^x dt \, e^{-t^2} \right )^2 = \left ( \int_0^{\infty} dt \, e^{-t^2} \right )^2 = \frac{\pi}{4} $$

Hint: In your expression for $F'$, try to write $x^2 (1+t^2)$ as $x^2 + (xt)^2$, use the law $e^{a+b} = e^a e^b$, move the constant term (that does not depend on $t$) outside of the integral, and change variables with $s = xt$.