How to compute $\int_{-1}^{1} e^{-1/(1-x^2)}dx$? As in the title, I would like to compute the integral:
\begin{equation}
\int_{-1}^{1}e^{-1/(1-x^2)}dx
\end{equation}
My hunch tells me that I should try to transform it to the correspoding $\int_{-1}^{1} e^{-y^2}dy$ but I am not sure. 
Thank you for your time!

$\int_{-1}^1e^{-\frac{1}{1-x^2}}~dx$
$=\int_{-1}^1e^{\frac{1}{x^2-1}}~dx$
$=\int_{-1}^0e^{\frac{1}{x^2-1}}~dx+\int_0^1e^{\frac{1}{x^2-1}}~dx$
$=\int_1^0e^{\frac{1}{(-x)^2-1}}~d(-x)+\int_0^1e^{\frac{1}{x^2-1}}~dx$
$=\int_0^1e^{\frac{1}{x^2-1}}~dx+\int_0^1e^{\frac{1}{x^2-1}}~dx$
$=2\int_0^1e^{\frac{1}{x^2-1}}~dx$
$=2\int_0^\infty e^{\frac{1}{\tanh^2x-1}}~d(\tanh x)$
$=2\int_0^\infty e^{-\frac{1}{\text{sech}^2x}}~d(\tanh x)$
$=2\int_0^\infty e^{-\cosh^2x}~d(\tanh x)$
$=2\left[e^{-\cosh^2x}\tanh x\right]_0^\infty-2\int_0^\infty\tanh x~d\left(e^{-\cosh^2x}\right)$
$=4\int_0^\infty e^{-\cosh^2x}\sinh x\cosh x\tanh x~dx$
$=4\int_0^\infty e^{-\cosh^2x}\sinh^2x~dx$
$=4\int_0^\infty e^{-\frac{\cosh2x+1}{2}}\dfrac{\cosh2x-1}{2}dx$
$=2e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh2x}{2}}(\cosh2x-1)~dx$
$=e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh2x}{2}}(\cosh2x-1)~d(2x)$
$=e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh x}{2}}(\cosh x-1)~dx$
$=e^{-\frac{1}{2}}\left(K_1\left(\dfrac{1}{2}\right)-K_0\left(\dfrac{1}{2}\right)\right)$
