<s>[INST] Can the integral $\int_0^\infty\ \frac{1}{\sqrt{x^2+a^2}} \cdot \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx$ be solved analytically? I'm trying to solve the following integrals:
$$I_1(a)=\int_0^\infty\ \frac{1}{\sqrt{x^2+a^2}} \cdot \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx$$
and
$$I_2(a)=\int_0^\infty\ \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx.$$
They seem similar to the two integrals solved in this post, so I wonder if there is something in the Bessel function family that will do the trick. But so far, substitutions like $x=a \sinh(t)$ and $u=\sqrt{x^2+a^2}$ are not giving me integrals that can be manipulated into one of the Bessel function forms on Wikipedia or Wolfram. 
 [/INST] Assume $a\neq0$ for the key case.
For $\int_0^\infty\dfrac{1}{\sqrt{x^2+a^2}\left(1+e^\sqrt{x^2+a^2}\right)}~dx$ ,
$\int_0^\infty\dfrac{1}{\sqrt{x^2+a^2}\left(1+e^\sqrt{x^2+a^2}\right)}~dx$
$=\int_0^\infty\dfrac{1}{\sqrt{a^2\sinh^2t+a^2}\left(1+e^\sqrt{a^2\sinh^2t+a^2}\right)}~d(|a|\sinh t)$
$=\int_0^\infty\dfrac{1}{1+e^{|a|\cosh t}}~dt$
$=\int_0^\infty\dfrac{e^{-|a|\cosh t}}{1+e^{-|a|\cosh t}}~dt$
$=\int_0^\infty\sum\limits_{n=0}^\infty(-1)^ne^{-|a|(n+1)\cosh t}~dt$
$=\sum\limits_{n=0}^\infty(-1)^nK_0(|a|(n+1))$
For $\int_0^\infty\dfrac{1}{1+e^\sqrt{x^2+a^2}}~dx$ ,
$\int_0^\infty\dfrac{1}{1+e^\sqrt{x^2+a^2}}~dx$
$=\int_0^\infty\dfrac{1}{1+e^\sqrt{a^2\sinh^2t+a^2}}~d(|a|\sinh t)$
$=\int_0^\infty\dfrac{|a|\cosh t}{1+e^{|a|\cosh t}}~dt$
$=\int_0^\infty\dfrac{|a|e^{-|a|\cosh t}\cosh t}{1+e^{-|a|\cosh t}}~dt$
$=\int_0^\infty\sum\limits_{n=0}^\infty|a|(-1)^ne^{-|a|(n+1)\cosh t}\cosh t~dt$
$=\sum\limits_{n=0}^\infty|a|(-1)^nK_1(|a|(n+1))$
</s>