Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.
The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express $\int x^2 \mathrm {d}x$ in elementary functions such as $\frac {x^3} {3} +C$. However, the indefinite integral from $ (-\infty,\infty)$ does exist and it is $\sqrt {\pi}$ so explicitly: $$\int^ {\infty}_ {-\infty} e^ {-x^2} = \sqrt {\pi}$$ Note ...
This is the integral that is sometimes described as "the area under the curve" (although I would consider that an application of the definite integral, not a definition).
A different type of integral, if you want to call it an integral, is a "path integral". These are actually defined by a "normal" integral (such as a Riemann integral), but path integrals do not seek to find the area under a curve. I think of them as finding a weighted, total displacement along a curve.
The integral of 0 is C, because the derivative of C is zero. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function f (x)=C will have a slope of zero at point on the function.
However, one "intrinsic integral closure" that is often used is the normalization, which in the case on an integral domain is the integral closure in its field of fractions. It's the maximal integral extension with the same fraction field as the original domain.