Question: How to evaluate the following definite integral:

$$\int\_{-1}^1 \exp\left(\frac1{x^2-1}\right) \, dx$$

It seems that indefinite integral also cannot be expressed in standard functions. I would like any solution in popular elementary or non-elementary functions.

Answer: 
I have the best approach:

$\int\_{-1}^1e^{\frac{1}{x^2-1}}~dx$

$=\int\_{-1}^0e^{\frac{1}{x^2-1}}~dx+\int\_0^1e^{\frac{1}{x^2-1}}~dx$

$=\int\_1^0e^{\frac{1}{(-x)^2-1}}~d(-x)+\int\_0^1e^{\frac{1}{x^2-1}}~dx$

$=\int\_0^1e^{\frac{1}{x^2-1}}~dx+\int\_0^1e^{\frac{1}{x^2-1}}~dx$

$=2\int\_0^1e^{\frac{1}{x^2-1}}~dx$

$=2\int\_0^\infty e^{\frac{1}{\tanh^2x-1}}~d(\tanh x)$

$=2\int\_0^\infty e^{-\frac{1}{\text{sech}^2x}}~d(\tanh x)$

$=2\int\_0^\infty e^{-\cosh^2x}~d(\tanh x)$

$=2\left[e^{-\cosh^2x}\tanh x\right]\_0^\infty-2\int\_0^\infty\tanh x~d\left(e^{-\cosh^2x}\right)$

$=4\int\_0^\infty e^{-\cosh^2x}\sinh x\cosh x\tanh x~dx$

$=4\int\_0^\infty e^{-\cosh^2x}\sinh^2x~dx$

$=4\int\_0^\infty e^{-\frac{\cosh2x+1}{2}}\dfrac{\cosh2x-1}{2}dx$

$=2e^{-\frac{1}{2}}\int\_0^\infty e^{-\frac{\cosh2x}{2}}(\cosh2x-1)~dx$

$=e^{-\frac{1}{2}}\int\_0^\infty e^{-\frac{\cosh2x}{2}}(\cosh2x-1)~d(2x)$

$=e^{-\frac{1}{2}}\int\_0^\infty e^{-\frac{\cosh x}{2}}(\cosh x-1)~dx$

$=e^{-\frac{1}{2}}\left(K\_1\left(\dfrac{1}{2}\right)-K\_0\left(\dfrac{1}{2}\right)\right)$
The symmetry of the integrand and variable substitution $t=1/(1-x^2)-1$ can be used to transform your integral as follows:
$$
I=\int\_{-1}^1 \exp\left(\frac{1}{x^2-1}\right)dx=\int\_0^1 2\exp\left(\frac{1}{x^2-1}\right)dx
=\frac{1}{e}\int\_0^{\infty}\frac{e^{-t}dt}{\sqrt{t}(1+t)^{3/2}}\,,
$$
Maple can evaluate this integral in terms of Meijer's $G$-function, just as obtained by J.M. after "coaxing" his Mathematica. Alternatively, this integral can be recognised as [Whittaker's function $W$](http://mathworld.wolfram.com/WhittakerFunction.html). The same conclusion can also be arrived at by noting that this integral is, effectively, a Laplace transform and using an appropriate command of Maple (or Mathematica), with the result
$$
I=\sqrt{\frac{\pi}{e}} W\_{-\frac{1}{2},-\frac{1}{2}}(1) \approx 0.44399\,.
$$
This answer is only slightly more elegant than J.M.'s; it's still not elementary and I am unsure whether you would describe Whittaker's function as "popular". You may also consider reformulating this in terms of confluent hypergeometric function.

**Updated 15/05/2012:** It seems that you can avoid using Whittaker's function after all, at the cost of computing a modified Bessel function and a certain continued fraction. Specifically, identities [13.17.3](http://dlmf.nist.gov/13.17.E3) and [13.18.9](http://dlmf.nist.gov/13.18.E9), given in the "new" DLMF, lead to the simple result:
$$
I=\frac{K\_0(1/2)}{C\sqrt{e}}\,,
$$
with constant $C$ given by the following continued fraction:
$$
C=1+\frac{1/2}{1+}\,\frac{3/2}{1+}\,\frac{3/2}{1+}\,\frac{5/2}{1+}\,\frac{5/2}{1+}\,
\frac{7/2}{1+}\,\frac{7/2}{1+}\,\cdots
=\frac{W\_{0,0}(1)}{W\_{-\frac{1}{2},-\frac{1}{2}}(1)}\approx 1.2628295456\,.
$$
In terms of functions involved, this is a lot more "elementary". From the computational point of view, I suspect that the original answer above is more practical.