integration - How do I Evaluate the Integral $int_{-infty }^{infty } frac{1}{e^{x^2}+1} , dx$?

Like the question states, how do I evaluate the integral
$$\int_{-\infty }^{\infty } \frac{1}{e^{x^2}+1} \, dx$$
I know of no methods to evaluate this, but when I plug this into Mathematica and Wolfram Alpha, it returns
$$\int_{-\infty }^{\infty } \frac{1}{e^{x^2}+1} \, dx=\left(1-\sqrt{2}\right) \sqrt{\pi } \zeta
\left(\frac{1}{2}\right)$$
where $\zeta (x)$ is the Riemann Zeta Function.

Answer

The given integral equals
$$ 2\int_{0}^{+\infty}\frac{dx}{1+e^{x^2}} = \int_{0}^{+\infty}\frac{dz}{\sqrt{z}(1+e^{z})}=\sum_{n\geq 1}(-1)^{n+1}\int_{0}^{+\infty}\frac{e^{-nz}}{\sqrt{z}}\,dz $$

or
$$ \sqrt{\pi}\sum_{n\geq 1}\frac{(-1)^{n+1}}{\sqrt{n}} = \sqrt{\pi}\,\eta\left(\frac{1}{2}\right)=(1-\sqrt{2})\sqrt{\pi}\,\zeta\left(\frac{1}{2}\right) $$
as claimed by WA.