Integral of $e^{-k \cosh(z)} \text {sech}(z) \ dz$ from $z=0$ to $z=\infty$

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I'm trying to evaluate $$ \int_0^1 \frac{e^{-k/(1-x^2)^{1/2}}}{(1-x^2)^{1/2}} dx $$

I attached an image with an attempt to solve with variable substitution and differentiation under the integral sign, but I couldn't finish it.

attempt

The problem is that the modified Bessel functions of second type and order 0 are infinite when alpha=0. We would then have a multiplication of zero by infinity. Is there an analytical way to find the constant C? Take a look and tell me if there is another better way to approach it. Note: L are modified Struve functions of order -1 and 0.

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For your curiosity.

A few years ago, I had to make, for fast calculations, some approximations of $$f(x)=L_{-1}(x)\,K_0(x)+L_0(x)\,K_1(x)$$ Using expansions around $x=0$, I obtained $$f(x)=\frac 1 \pi \sum_{n=0}^\infty \frac {a_n-b_n\,t}{c_n} x^{2n}\quad \text{with} \quad t=\log \left(\frac{x}{2}\right)+\gamma$$ the first coefficients being $$\left( \begin{array}{cccc} n & a_n & b_n & c_n \\ 0 & 2 & 2 & 1 \\ 1 & 4 & 3 & 18 \\ 2 & 17 & 10 & 1600 \\ 3 & 83 & 42 & 338688 \\ 4 & 79 & 36 & 23887872 \end{array} \right)$$

Using the above term, for $0 < x \leq 3$, the infinite norm is $1.77 \times 10^{-8}$ and the maximum error is $3.42\times 10^{-4}$.