"an indefinite integral is valid only on a interval"?

Question

Ind. Int.="Indefinite Integral"
In my book (Stewart) they say

"Recall that the most general antiderivative on a given interval is
obtained by adding a constant to a particular antiderivative. [Bold]We adopt the convention that
when a formula for a general Ind. Int. is given, it is valid only on an interval.[/Bold] Thus, we write

Ind. Int. 1/x^2 dx = - (1/x) + C

with the understanding that it is valid on the interval (0, infinite) or on the interval (-infinite, 0). This is true despite the fact that the general antiderivative of the function , f(x)=1/x^2, x ≠ 0 , is:

F(x):
- (1/x) + C1 if x<0
- (1/x) + C2 if x>0"

Well, I don't understand this convention, don't know if it is something too obvious and I'm complicating myself, like, "there could be points where the Ind. Int. isn't defined" or maybe there's a subtle point behind this, maybe they want to say that a Ind. Int. that holds for any interval, no matter how small, it's a valid Ind. Int of the function.
Thank u.

Answer 1

"there's a subtle point behind this, maybe they want to say that a Ind. Int. that holds for any interval, no matter how small, it's a valid Ind. Int of the function."
no, that's not true
They only say that is a convention of writing
they write: "Ind. Int. 1/x^2 dx = - (1/x) + C" because it is shorter and easier to manipulate.
They say that the real value of the integral is :
" F(x):
- (1/x) + C1 if x<0
- (1/x) + C2 if x>0"
but they prefer not to use this notation, even if it true.

, "there could be points where the Ind. Int. isn't defined"
yes, that is true
The indefinite integral is valid on some interval. But they prefer not to write the integral to make the computation easier.
Imagine how you would write:
ind int 1/x^2 + ind int 1/(x-1)^2 if we wouldn't have this convention
It will be -(1/x) + -(1/x-1) +C1+D1, if x<0
-(1/x) + -(1/x-1) +C2+D1, if 0 -(1/x) + -(1/x-1) +C2+D2, if x>1

while if we use the convention, the sum will be only
-(1/x) + -(1/x-1) +C+D,
Mathematics is governed by exact rules. In this case, you have the rule of "not being exact for awhile", but is still a rule.

Answer 2

This occurs ony if the antiderivative has a discontinuity. In your case, there is a discontinuity at x= 0, because division by zero is undefined.

It is clear that the ind int exists everywhere except at x = 0, and hence it is common to have this to be understood just as we understand that 5 is positive without writing a + sign in front of it.

If the derivative were x^2+x, there would be np problem as the ind .int is continuous everywhere.