I will try to explain my question throughly as possible.
According to the wikipedia, it explains the fundamental theorem of calculus as:
Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by
$F(x) = \int_a^x f(t)dt$.
Then, F is uniformly continuous on [a, b], differentiable on the open interval (a, b), and
$F'(x) = f(x)$ for all x in (a,b).
So for example, if we have the function $f(t) = \frac{1}{t^2-1}$. If the domain of $x$ is $(-\infty,\infty)$. My thought is to use this theorem I have consider when $(-\infty, a]$ and $[a, \infty)$. So for example. If $a$ is 1, then
$F(x) = \int_1^x f(t)dt$.
Hence need to consider
$F(\infty) = \int_1^{\infty} f(t)dt$ and $F(-\infty) = \int_1^{-\infty} f(t)dt$. Since the second one is invalid, then $F(-\infty) = -\int_{-\infty}^{1} f(t)dt$.
Using this, can we say $F'(x) = f(x)$ for all x in $(-\infty, 1)$ and $(1, \infty)$?
an integral is only defined on $[a, b]$ as $a<b<\infty$. for any function f, $f(\infty)$ is not even to be defined. what you can ask instead is on $$lim_{x\rightarrow\infty} f(x)$$ does it exist? infinite? when looking on Improper integral defined on $(a, \infty)$ as $a <\infty$, this integral can only be discussed on if $$lim_{x\rightarrow\infty} \intop_{a}^{x} f < \infty$$ but even before that, more significantly - does this limit even exist. further more, for on the two diffrent intrvals, $(a,\infty), (-\infty, a)$ you can say $F'(x) = f(x)$ if and only if $lim_{x\rightarrow\infty} \intop_{a}^{x} f < \infty$ and $lim_{x\rightarrow-\infty} \intop_{x}^{a} f < \infty$ both are finite and same sign (both must be non-negetive $(\leq)$ or negative). in ant other terms, the answer is defiantly no.