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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=e^{-\frac{1}{2}}\left(K\_1\left(\dfrac{1}{2}\right)-K\_0\left(\dfrac{1}{2}\right)\right)$
Using the symmetry $x \to -x$ and change of variables $t = 1/\sqrt{1-x^2}$ we get
$2 \int\_1^\infty \dfrac{e^{-t^2}}{t^2 \sqrt{t^2-1}} dt$, which Maple 16 can then evaluate as
${{\rm e}^{-1/2}} \left( {{\rm K}\_1\left(1/2\right)}-
{{\rm K}\_0\left(1/2\right)} \right) $.