Methods to solve $\int_{0}^{\infty} \frac{e^{-x^2}}{x^2 + 1}\:dx$

You can use Plancherel's theorem. Note that $$ 2I = \int_{-\infty}^{\infty} \frac{e^{-x^2}}{x^2 + 1}dx. $$Let $f(x) = e^{-x^2}$ and $g(x) = \frac{1}{1+x^2}$. Then we have $$ \widehat{f}(\xi) = \sqrt{\pi}e^{-\pi^2\xi^2}, $$ and $$ \widehat{g}(\xi) = \pi e^{-2\pi|\xi|}. $$ By Plancherel's theorem, we have $$\begin{eqnarray} \int_{-\infty}^{\infty} f(x)g(x)dx&=&\int_{-\infty}^{\infty} \widehat{f}(\xi)\widehat{g}(\xi)d\xi\\&=&\pi^{\frac{3}{2}}\int_{-\infty}^{\infty}e^{-\pi^2\xi^2-2\pi|\xi|}d\xi\\ &=&2\pi^{\frac{3}{2}}\int_{0}^{\infty}e^{-\pi^2\xi^2-2\pi\xi}d\xi\\ &=&2\pi^{\frac{3}{2}}e\int_{\frac{1}{\pi}}^{\infty}e^{-\pi^2\xi^2}d\xi\\ &=&2\pi^{\frac{1}{2}}e\int_{1}^{\infty}e^{-\xi^2}d\xi = \pi e \operatorname{erfc}(1). \end{eqnarray}$$ This gives $I = \frac{\pi}{2}e \operatorname{erfc}(1).$

Here is a method that employs the old trick of converting the integral into a double integral.

Observe that $$\frac{1}{1 + x^2} = \int_0^\infty e^{-u(1 + x^2)} \, du.$$ So your integral can be rewritten as $$I = \int_0^\infty e^{-x^2} \int_0^\infty e^{-u(1 + x^2)} \, du \, dx.$$ or $$I = \int_0^\infty e^{-u} \int_0^\infty e^{-(1 + u)x^2} \, dx \, du,$$ on changing the order of integration.

Enforcing a substitution of $x \mapsto x/\sqrt{1 + u}$ gives $$I = \int_0^\infty \frac{e^{-u}}{\sqrt{1 + u}} \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \int_0^\infty \frac{e^{-u}}{\sqrt{1 + u}} \, du.$$

Next exforcing a substitution of $u \mapsto u^2 - 1$ gives $$I = \sqrt{\pi} e \int_1^\infty e^{-u^2} \, du = \sqrt{\pi} e \cdot \frac{\sqrt{\pi}}{2} \text{erfc} (1) = \frac{\pi e}{2} \text{erfc} (1),$$ as expected.