Inverse Laplace Transform
First you have to remember a Laplace transform property:

$$ e^{a t} u(t) \longleftrightarrow \frac{1}{s-a} ~~~,~~~ \mathcal{Re}\{s\} &gt; \mathcal{Re}\{a\}   \tag{1} $$

$$ -e^{a t} u(-t) \longleftrightarrow \frac{1}{s-a} ~~~,~~~ \mathcal{Re}\{s\} &lt; \mathcal{Re}\{a\}   \tag{2} $$

This property states that a given Laplace transform $H(s)$ can correspond to multiple inverse transforms, depending on the region of convergence ROC. 

Therefore given $H(s) = \frac{1}{s-a}$ , you can find two possible inverses as $x(t) = e^{at} u(t)$ or $x(t) = -e^{at} u(-t)$ depending on whether the ROC is to the left or right of the pole location. Note that one of them is causal and the complementary is anti-causal.

Now, given your transfer function 

$$ H(s) = \frac{s-1}{(s+1)(s-2)} = \frac{2/3}{s+1} + \frac{1/3}{s-2} ,  \tag{3} $$ 

it has two poles at $s = -1$ and $s=2$. There will be three possible ROC's with three different inverses :

$$ 
\begin{align}
\mathcal{Re}\{s\} &lt; \mathcal{Re}\{-1\} &amp; \implies h(t) = -\frac{2}{3} e^{-t}u(-t) - \frac{1}{3} e^{2t} u(-t)  \tag{4} \\
\mathcal{Re}\{-1\} &lt; \mathcal{Re}\{s\} &lt; \mathcal{Re}\{2\}  &amp;\implies h(t) = \frac{2}{3} e^{-t}u(t) - \frac{1}{3} e^{2t} u(-t)  \tag{5} \\
\mathcal{Re}\{s\} &gt; \mathcal{Re}\{2\} &amp; \implies h(t) = \frac{2}{3} e^{-t}u(t) + \frac{1}{3} e^{2t} u(t)  \tag{6}
\end{align}
$$

The impulse response in (4) is anti-causal and unstable.

The impulse response in (5) is non-causal and stable.

The impulse response in (6) is causal and unstable.