How to find $\int \frac{e^{-x^2}}{x^2 + 1} dx$?

$$I_n=\int_{-n}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}\,dx=2\int_{0}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}\,dx$$ Let $$nx^2=t\implies x=\frac{\sqrt{t}}{\sqrt{n}}\implies dx=\frac{dt}{2 \sqrt{n} \sqrt{t}}$$ making $$I_n=\sqrt{n}\int_0^1\frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt$$ You do not need to compute anything else to show that $I_n$ is just proportional to $\sqrt{n}$ and just conclude.

Now, as said in answers, the last integral cannot be computed and series expansions are required. Using Taylor, we would have $$\frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}=\frac{1}{\sqrt{t}}-\frac{3 \sqrt{t}}{2}+\frac{5 t^{3/2}}{3}-\frac{41 t^{5/2}}{24}+\frac{103 t^{7/2}}{60}-\frac{1237 t^{9/2}}{720}+O\left(t^{11/2}\right)$$ $$\int \frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt=2 \sqrt{t}-t^{3/2}+\frac{2 t^{5/2}}{3}-\frac{41 t^{7/2}}{84}+\frac{103 t^{9/2}}{270}-\frac{1237 t^{11/2}}{3960}+O\left(t^{13/2}\right)$$ Using the bounds, we should get $$\int \frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt=\frac{1634621}{1081080}\approx 1.51203$$ while numerical integration would lead to $\approx 1.38990$