Is there a way to tell whether an integral exists or not without graphing the function? For example, trying to integrate the function 1/(t-4)^2 from x=1 to x=5. If you just go through the motions and apply u-substitution and part two of the fundamental theorem, you can arrive at an answer for this just fine. But it was only after graphing it that I saw that there was an asymptote within the limits of the integral, and so obviously the integral would be undefined.
Is there any way to tell whether this will be the case at first glance?
Is the only way to just get good at recognizing when functions have an asymptote? Isn't this sometimes very difficult to predict without fulling graphing the function? 
Any help is appreciated.
Edit: one thing that should've been obvious to me is that: the function will have an asymptote if there's an opportunity for division by zero. So now my question is: is there anything else that can cause asymptotic behavior, and are there any situations in which integrals are undefined other than when the graph has an asymptote? 
 A: Well, it is not true that the answer is not defined; consider $\int_0^1 x^{-1/2}\,dx$.
Generally speaking, do not rely on purely graphical methods to determine anything! Yes, the presence of an asymptote should make you pause, and carefully check something, but pictures are lies as my analysis professor used to say.
As far as recognizing when a function will have a singularity, really you are just hunting for the possibility of dividing by zero; with enough time and patience you will find that you do not need to graph anything at all to recognize that.
EDIT: To add more information based on conversations and updates to the OP, let me discuss two phenomena that can also cause an integral to fail to have a value.
First, we turn our attention to the indicator function of the rationals:
$$
\chi_{\mathbb{Q}}(x) = \left\{
        \begin{array}{ll}
            1 & \quad x \in\mathbb{Q} \\
            0 & \quad x \not\in\mathbb{Q}
        \end{array}
    \right.
$$
This function, if you somehow were able to graph it, would look like two lines -- the line $y=1$ and the line $y=0$. But it isn't! The rationals are what we call "dense" in the reals, and so in reality there is just a very fine "cloud" of points, everywhere, that is hovering at $1$, and another cloud of points down at $0$. If you try and apply your Darboux or Riemann integral definition to this function on any interval $[a,b]$, you will find that the limits/sups/infs/whatevers don't agree, and thus the integral doesn't exist.
Second, we consider improper integrals over oscillatory functions. Consider the integral $\int_0^\infty \sin(2\pi x)\,dx$. The value (if it were to exist) of this integral would be $\lim_{b\rightarrow\infty}1-\cos(2\pi b)/2\pi$. But this limit doesn't exist, because $\cos(2\pi x)$ just "wiggles" off into infinity. Even though there doesn't appear to be a singularity in this function, there is; it has been obfuscated and pushed out to the upper limit of integration! 
