Finding Inverse Laplace Transform Calculate the inverse Laplace transform of: $$\frac{1}{s\cdot(\sqrt{s}+1)} \cdot e^{-\sqrt{s} \cdot x}$$
My attempt at the solution was to break down the fraction with partial fraction decomposition as follows: $$\frac{1}{s \cdot (\sqrt{s}+1)} = \frac1s-\frac{1}{\sqrt{s}}+\frac{1}{1+\sqrt{s}}$$
Then the first part can be easily computed from the table or by using some software: $$\mathcal{L}^{-1}( s^{-1} \cdot e^{-\sqrt{s} \cdot x} )= 1-erf(\frac{x}{2\sqrt{t}})$$
However the second part is not at all trivial, as I was unable to find any coherent answer to the problem: $$\mathcal{L}^{-1}( (-\frac{1}{\sqrt{s}}+\frac{1}{1+\sqrt{s}}) \cdot e^{-\sqrt{s} \cdot x} )$$
Could someone, please, guide towards the correct answer?
I have tried using computer algebra systems such Mathematica, but nothing seems to work.
***Treat x as a constant with x>0.

 A: $$\color{red}{\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{\left(1+\sqrt{s}\right) s}\right](t)}=\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s}
   x}}{1+\sqrt{s}}+\frac{e^{-\sqrt{s} x}}{s}-\frac{e^{-\sqrt{s} x}}{\sqrt{s}}\right](t)=\color{red}{\text{erfc}\left(\frac{x}{2
   \sqrt{t}}\right)-e^{t+x} \text{erfc}\left(\frac{2 t+x}{2 \sqrt{t}}\right)}$$
With CAS help:
$$\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{s}-\frac{e^{-\sqrt{s} x}}{\sqrt{s}}\right](t)=-\frac{e^{-\frac{x^2}{4 t}}}{\sqrt{\pi }
   \sqrt{t}}+\text{erfc}\left(\frac{x}{2 \sqrt{t}}\right)$$ 
It's hard to find $\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{1+\sqrt{s}}\right](t)$?
$$\mathcal{L}_s^{-1}\left[\frac{e^{-\sqrt{s} x}}{1+\sqrt{s}}\right](t)=\mathcal{L}_s^{-1}\left[\mathcal{L}_a^{-1}\left[\frac{e^{-\sqrt{s}
   x}}{a+\sqrt{s}}\right](q)\right](t)=\mathcal{L}_q\left[\mathcal{L}_s^{-1}\left[e^{-q \sqrt{s}-\sqrt{s}
   x}\right](t)\right](1)=\mathcal{L}_q\left[\frac{e^{-\frac{(q+x)^2}{4 t}} (q+x)}{2 \sqrt{\pi }
   t^{3/2}}\right](1)=\frac{e^{-\frac{x^2}{4 t}}}{\sqrt{\pi } \sqrt{t}}-e^{t+x} \text{erfc}\left(\frac{2 t+x}{2 \sqrt{t}}\right)$$
