Brian's suggestion above deserves a mention as I was able to solve the integral that way too!
Let $$I(t) = \int_{-\infty}^{\infty} \frac{\exp(-tx^2)}{1+x^4}dx$$ for a parameter $t \geq 0$. Differentiating gives
$$I'(t) = -\int_{-\infty}^{\infty} \frac{x^2 \exp(-tx^2)}{1+x^4}dx$$ and a second differentiation gives $$I''(t) = \int_{-\infty}^{\infty} \frac{x^4 \exp(-tx^2)}{1+x^4}dx.$$
Hence,
$$I''(t) + I(t) = \int_{-\infty}^{\infty} \exp(-tx^2) = \sqrt{\frac{\pi}{t}}$$
as Brian pointed out.
Recalling my lecture notes, we now have a second order inhomogenous linear ODE with constant coefficients. The solution can be written
$$I(t) = I_C(t) + I_P(t)$$ where $I_C(t)$ solves the homogenous ODE
$$I_C''(t) + I_C(t) = 0.$$
Let $I_1(t) = \sin(t)$ and $I_2(t) = \cos(t)$ be the two solutions to the homogenous ODE. The particular solution $I_P(t)$ can now be found using the method described here http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx
Noting that the Wronskian for I1 and I2 is -1, the particular solution is
$$I_P(t) = \sin(t) \int_{0}^t \sqrt{\frac{\pi}{u}} \cos(u) du - \cos(t) \int_{0}^t \sqrt{\frac{\pi}{u}} \sin(u) du$$
or
$$I_P(t) = 2\sqrt{\pi} \left( \sin(t) \int_{0}^{\sqrt{t}} \cos(u^2) du - \cos(t) \int_{0}^{\sqrt{t}} \sin(u^2) du \right).$$
Putting this all together gives
$$I(t) = A \sin(t) + B \cos(t) + \pi \sqrt{2} \left\{ C\left( \sqrt{\frac{2t}{\pi}} \right) \sin(t) - S\left( \sqrt{\frac{2t}{\pi}} \right) \cos(t) \right\},$$
where $C(x)$ and $S(x)$ are the Fresnel cosine and sine integrals respectively.
The initial conditions are
$$I(0) = \int_{-\infty}^{\infty} \frac{dx}{1+x^4}$$
and
$$-I'(0) = \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx.$$
Using a D shaped contour in the complex plane, these integrals can be shown to both equal $\frac{\pi}{\sqrt{2}}$. Hence $B = -A = \frac{\pi}{\sqrt{2}}$.
This finally gives
$$I(1) = \pi\cos(1) \frac{1 - 2 S\left( \sqrt{\frac{2}{\pi}} \right)}{\sqrt{2}} - \pi\sin(1) \frac{1 - 2 C\left( \sqrt{\frac{2}{\pi}} \right)}{\sqrt{2}}.$$