What is "improper" about improper integrals of type 2? What is the technical reason that one cannot integrate over vertical asymptotes?
I know integrals like $\int_{-1}^1-\frac{1}{x^2}dx$ are improper integrals of type 2 and that one should divide the domain in $[-1,0)$ and $(0,1]$.
But why exactly? What is it about the nature of integration that it wouldn't work to directly integrate over $[-1,1]$?
 A: Edit: Please notice the comment below this answer. I realized that the word "continuous" is not accurate enough in the text below. With some research I realized that the topic of integration for non-continuous and negative functions is too complex for me (e.g. Lebesgue's Theory of Integration: Its Origins and Development), and that I was not aware of what is involved. Having said that, I will keep the answer un-deleted in case some one benefits from the references and the comment below.
The problem is that the function is not continuous for all points in the range [-1, 1]. We have to break the integral range to 2 intervals first. The point of discontinuity here is obviously.
Here is a good short video:
Improper Integral of type II
Also, the answer here may be of further interest:
Does the fundamental theorem of calculus require continuity of the function being integrated?

A: If you directly integrate $$\int _{-1}^1 \frac {-1}{x^2} dx$$ you come up with the answer $$\int _{-1}^1 \frac {-1}{x^2}dx=2$$
When you sketch the graph you will see the graph is below the $x$-axis, so obviously the answer is wrong.
A: The Riemann integral itself has this property:
$$
\int_{-1}^1\frac{dx}{x^2} = \int_{-1}^0\frac{dx}{x^2} + \int_{0}^1\frac{dx}{x^2}.\tag{$1$}
$$
But this integral is not Riemann integrable, and ($1$) fails.  We cannot compute $(-\infty)+(+\infty)$ to get $0$.  As far as the theory of the Riemann integral is concerned, this integral does not exist.  So it has to be done in some other way.  The standard way to do it is the "improper integral" you mention.
A second way, which lacks important properties such as ($1$), but is preferred by physicists, is the "princpal value":
$$
\text{PV}\int_{-1}^1\frac{dx}{x^2} = \lim_{\delta \to 0^+}\left(\int_{-1}^{-\delta}\frac{dx}{x^2} + \int_{\delta}^1\frac{dx}{x^2}\right).\tag{$2$}
$$ 
Thirdly (I claim) the Lebesgue integral is superior to the Riemann integral for this  purpose in that it allows computation of integrals for unbounded functions, and for functions undefined on a set of measure zero.
