$\int_{-\infty}^\infty \frac{e^{-\lvert x \rvert}}{x^4+1}dx$ $$\int_{-\infty}^\infty \frac{e^{-\lvert x \rvert}}{x^4+1}dx$$
Wolframalpha says the answer is 1.26096, which is not helpful.
I tried with contour integral, but I should divide $x > 0$ and $x < 0$ domain and I'm not sure how to proceed from there, I mean, don't know how to draw the contour. I don't think I can draw contour through imaginary axis cuz the integral would diverge there.
 A: I know not of nice closed form, but one has
$$\frac1{x^4+1}=\frac14\sum_{\omega^4=-1}\frac\omega{x+\omega}$$
and
$$\int_0^\infty\frac{e^{-x}}{x+\omega}~dx=e^\omega E_1(\omega)$$
where $E_1$ is an exponential integral. Hence, your integral is given by

$$\int_{-\infty}^\infty\frac{e^{-|x|}}{x^4+1}~dx=\frac12\sum_{\omega^4=-1}\omega e^\omega E_1(\omega)$$

A: Just an addendum to SBA's answer.
$$ \int_{\mathbb{R}}\frac{e^{-|x|}}{x^4+1}\,dx = 2\sqrt{2}\int_{\mathbb{R}}\frac{e^{-|x|/\sqrt{2}}}{(x^2+2x+2)(x^2-2x+2)}\,dx $$
boils down to
$$ \frac{1}{2\sqrt{2}}\int_{\mathbb{R}}e^{-|x|/\sqrt{2}}\left(\frac{2-x}{x^2-2x+2}+\frac{2+x}{x^2+2x+2}\right)\,dx $$
then to (by parity)
$$ \frac{1}{\sqrt{2}}\int_{0}^{+\infty}e^{-x/\sqrt{2}}\left(\frac{2-x}{x^2-2x+2}+\frac{2+x}{x^2+2x+2}\right)\,dx. $$
Through the Fourier transform the original integral can be written as 
$$ I=2\int_{0}^{+\infty}\frac{\cos(s)+\sin(s)}{e^s(1+2s^2)}\,ds $$
and easily approximated through the Cauchy-Schwarz inequality ($I^2\leq\frac{3\pi}{4\sqrt{2}}$), since the new integrand function has a Gaussian behaviour in a neighbourhood of the origin, where most of the mass of the integral is concentrated. A not terrible approximation is provided by 
$$ I \approx 2\int_{0}^{+\infty}\frac{ds}{\left(1+\frac{3}{2}s^2\right)^2} = \frac{\pi}{\sqrt{6}}. $$
I am not expecting a closed form beside the one provided by the exponential integral.
