Find the domain $J$ of $F$ and compute, if existing, $F(0)$, where $$F(x)=\int_{-\infty}^xf(t)\mathrm dt, \qquad f(t)=\begin{cases}h(t)=\dfrac1{(t-1)(t-2)}&t<0 \\ \\ g(t)=\sqrt[3]{\left(\dfrac{t+1}{t}\right)^2}&t\ge0\end{cases}$$
I computed $$\operatorname{Dom}h=\mathbb R\setminus\{1,2\}\cap (-\infty,0)=(-\infty,0)\\ \operatorname{Dom}g=\mathbb R\setminus\{0\}\cap [0,+\infty)=(0,+\infty)$$
Then $(-\infty,0)\subseteq J$.
$$\lim_{x\to0^-}h(x)=\frac12\implies 0\notin J.$$
So $J=(-\infty,0)$ and $\nexists F(0)$. Am I right?
The domain of $h$ is ${\mathbb R}\setminus\{1,2\}$, and the domain of $g$ is ${\mathbb R}\setminus\{0\}$. Strictly speaking the function $f$ is undefined at $t=0$. But you can make sense of the function $$F(x):=\int_{-\infty}^x f(t)\>dt\qquad(-\infty<x<\infty)$$ nevertheless. In other words: $J={\mathbb R}$. This is true without much ado in the Lebesgue world, and can be secured under the Riemann integral as follows:
When $x<0$ we have to consider the improper integral $$F(x):=\lim_{a\to-\infty}\int_a^x h(t)\>dt\ .$$ One has $$\int_a^x h(t)\>dt=\int_a^x\left({1\over1-t}-{1\over2-t}\right)\>dt=\log{2-t\over1-t}\Biggr|_a^x\ .$$ It follows that $$F(x)=\log{2-x\over1-x}\qquad(x<0)\ ,$$and $F(0)=\lim_{x\to0-} F(x)=\log 2$.
Now for $x\geq0$ we note that $$0\leq g(t)\leq 2 t^{-2/3}\qquad(0\leq t\leq1)\ .$$ It follows that the improper integral $$\int_0^x g(t)\>dt:=\lim_{a\to0+}\int_a^x g(t)\>dx$$ exists (I don't bother computing it). We therefore can say that $$F(x)=\log 2+\int_0^x g(t)\>dt\qquad(x\geq0)\ ,$$ and our $F$ is even continuous on all of ${\mathbb R}$.