Inverse Laplace of $ \frac{1}{\sqrt{s} - 1} $? please help with this.
I found this in textbook.
Not derived from any differential equation.
Also found the answer $$ \frac{1}{\sqrt{\pi}\sqrt{t}} + e^t * erf(\sqrt{t}) $$
(but don't know how) 
 A: Start from Babak's observation that
$$\frac{1}{\sqrt{s}-1} = \frac{\sqrt{s}}{s-1} + \frac{1}{s-1}$$
The ILT of the second term is simply $e^t$.  For the first term, consider the following integral:
$$\oint_C dz \,e^{z t} \frac{\sqrt{z}}{z-1}$$
where $C$ is as follows:

We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis.  There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.
$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.
$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.
$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.
$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.
$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.
$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.
We will show that the integral along $C_2$,$C_4$, and $C_6$ vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.
On $C_2$, the real part of the argument of the exponential is
$$R t \cos{\theta} $$
where $\theta \in [\pi/2,\pi)$.  Clearly, $\cos{\theta} < 0$, so that the integrand exponentially decays as $R \rightarrow \infty$ and therefore the integral vanishes along $C_2$.
On $C_6$, we have the same thing, but now $\theta \in (-\pi,-\pi/2]$.  This means that, due to the evenness of cosine, the integrand exponentially decays again as $R \rightarrow \infty$ and therefore the integral also vanishes along $C_6$.
On $C_4$, the integral vanishes as $\epsilon$ in the limit $\epsilon \rightarrow 0$.  Thus, we are left with the following by the residue theorem:
$$\left [ \int_{C_1} + \int_{C_3} + \int_{C_5}\right] dz \: \frac{\sqrt{z}}{z-1} e^{z t} = i 2 \pi \, e^t$$
because of the pole at $z=1$.
On $C_3$, we parametrize by $z=e^{i \pi} x$ and the integral along $C_3$ becomes
$$\int_{C_3} dz \: \frac{\sqrt{z}}{z-1} e^{z t} = \int_{\infty}^0 dx \: \frac{i \sqrt{x}}{x+1} e^{-x t}$$
On $C_5$, however, we parametrize by $z=e^{-i \pi} x$ and the integral along $C_5$ becomes
$$\int_{C_5} dz \: \frac{\sqrt{z}}{z-1} e^{z t} = \int_0^{\infty} dx \: \frac{-i \sqrt{x}}{x+1} e^{-x t}$$
We may now write
$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \:\frac{\sqrt{s}}{s-1} e^{s t} = e^t + \frac{1}{\pi} \int_0^{\infty} dx \frac{\sqrt{x}}{x+1} e^{-x t}$$
Now consider the integral on the RHS:
$$ \begin{align}\int_0^{\infty} dx \frac{\sqrt{x}}{x+1} e^{-x t} &= \int_{-\infty}^{\infty} du \frac{u^2}{1+u^2} e^{-t u^2}\\ &=\int_{-\infty}^{\infty} du e^{-t u^2} -\int_{-\infty}^{\infty} du \frac{1}{1+u^2} e^{-t u^2}\\ &= \sqrt{\frac{\pi}{t}} - \pi \,e^t \, \text{erfc}{\sqrt{t}}\end{align}$$
Putting this all together, we get
$$\frac{1}{\sqrt{\pi t}} + e^t \text{erf}{\sqrt{t}} + e^t = \frac{1}{\sqrt{\pi t}} + e^t  (1+\text{erf}{\sqrt{t}})$$
Your answer is missing an $e^t$.
A: $$\frac{1}{\sqrt{s}-1}=\frac{\sqrt{s}+1}{s-1}=\frac{\sqrt{s}}{s-1}+\frac{1}{s-1}=\frac{s}{\sqrt{s}(s-1)}+\frac{1}{s-1}=\frac{1}{\sqrt{s}(s-1)}+\frac{1}{\sqrt{s}}+\frac{1}{s-1}$$
 And we know that $$\mathcal{L}(\text{erf}(\sqrt{t}))=\frac{1}{s\sqrt{s+1}},~\mathcal{L}(e^{at}f(t))=\mathcal{L}(f(t))|_{s\to s-1},~\mathcal{L}\left(\frac{1}{\sqrt{t}}\right)=\sqrt{\frac{\pi}{s}}$$
