Why doesn't $\int_{-1}^{1}\frac{dx}{x} = \ln|x|\biggr\rvert_{-1}^{1} = 0$? $1/x$ is an odd function, so it makes sense to me intuitively that the area would be $0$, and similarly I would expect that $\int_{-1}^{2}\frac{dx}{x} = \ln(2)$.
Proof Wiki seems to confirm my intuition, but with the exception of functions that don't have a primitive (i.e. integral?), which I guess this one doesn't, because of the discontinuity at $x=0$.
Nonetheless, it seems to me that the area under $1/x$ must be $0$ because:
$$\int_{-1}^{1}\frac{dx}{x} = \lim_{a\to0} \left[ \int_{-1}^{a}{x^{-1} + \int_{a}^{1}{x^{-1}}} \right] = 0$$
I just can't shake the intuitive feeling that the area is $0$. Bonus points if you can explain why it is not $0$ in an intuitive way.
 A: Your intuition (expressed in the comments below the OP) is that the integral in question is analogous to $$\lim_{x \to \infty}(x-x)$$
which is, of course, $0$.  But a better way to think about it is to consider
$$\lim_{(x,y)\to(\infty,\infty)}(x-y)$$
which informally can be thought of as $\infty - \infty$, but this cannot be simply expressed as $0$.  More precisely, the value of that limit depends on how $x$ and $y$ go to infinity -- what path they follow, what rate they diverge relative to one another.
Likewise, when you look at an integral of an odd function over a symmetric interval with a pole in it, it may be very tempting to imagine approaching the vertical asymptote from the left and right sides at the same "rate"; if you do that then the areas always cancel out, and you end up with $0$.  This is what other answers have referred to as the Cauchy principal value. But you could also take those two limits (approaching from the left and from the right) independently, at different rates, in which case the areas would not cancel, and almost anything is possible.
A: Simple: $\int_{-1}^{1} \frac{dx}{x}$ is not well defined. The area, if it should exist in whatever definition of it you choose, is certainly not given by this expression.
A: 
The integral equals zero if you adopt the Cauchy principal value concept. 

A: The problem here is that, if you integrate on an interval, then the signed area is well-defined, because the shape is bounded. But an unbounded shape, such as the region of the plane enclosed by $y = 1/x, x = -1, x = 1, y = 0$, cannot have well-defined signed area. And while $1 - 1 = 0$, and $100000000000 - 100000000000 = 0$, $\text{Undefined} - \text{Undefined}$ is not $0$. It is true that shape in question has reflective symmetry with respect to the origin, but symmetry only makes the signed area $0$ if the signed area exists in the first place. Our intuition forgets this because we typically work with bounded shapes, and our brains have the very bad and incorrect habit of assuming that things that hold for the bounded (finite) also hold for the unbounded (infinite), even though this is not the case.
