How to solve $\int_{0}^{\infty} e^{-ax} \frac{x}{\sqrt{1+x^2}}dx$ or $\int_{0}^{\infty} e^{-ax} \sqrt{1+x^2}dx$ I need to solve either of the following integrals, they are related to each other by integration by parts:

$$I_1 = \int_{0}^{\infty} e^{-ax} \frac{x}{\sqrt{1+x^2}}dx$$

or

$$I_2 = \int_{0}^{\infty} e^{-ax} \sqrt{1+x^2}dx$$

I have not been able to find them in integral tables and I wonder if it is possible to solve either of these integrals.

Any ideas on how to calculate $I_1$ or $I_2$?
For $\int_0^\infty e^{-ax}\dfrac{x}{\sqrt{1+x^2}}~dx$ ,

$\int_0^\infty e^{-ax}\dfrac{x}{\sqrt{1+x^2}}~dx$

$=\int_0^\infty e^{-a\sinh t}\dfrac{\sinh t}{\sqrt{1+\sinh^2t}}~d(\sinh t)$

$=\int_0^\infty e^{-a\sinh t}\sinh t~dt$

$=-\dfrac{d}{da}\int_0^\infty e^{-a\sinh t}~dt$

$=-\dfrac{\pi}{2}\dfrac{d}{da}\mathbf{K}_0(a)$ (according to https://dlmf.nist.gov/11.5)

For $\int_0^\infty e^{-ax}\sqrt{1+x^2}~dx$ ,

$\int_0^\infty e^{-ax}\sqrt{1+x^2}~dx=\dfrac{\pi\mathbf{K}_1(a)}{2a}$ (according to https://dlmf.nist.gov/11.5)
If you notice it, you will see that this is the form of Laplace transformation. Thus the first integration can be written as the Laplace transformation of (x/root(x^2 +1))
The second integral can be written by integration by parts
Which is e^-ax * root(x^2 +1) - integrate(-ae^-at * root(x^2 +1) dx)
Which is then equals to 
e^-ax * root(x^2 +1) + a*laplace of (root(x^2 +1))

Hope this helped