How treat a point where a function is not defined for indefinite integral I have a question maybe stupid but I don't know how solve it.
Solving the indefinite integral I have a doubt:
When I have a definite integral usually we check if  the interval of integration is finite or if the function to be integrated is well defined, in the interval of integration, if not I am talking of a improper integral.
For example $\int_0^x \frac{1}{x}$ is an improper integral since $\lim_{x\to 0^+}\frac{1}{x}=+\infty$.
$\textbf{My question:}$ when I have an indefinite integral, for instance $\int \frac{1}{x} dx$, and for $x=0$ the function $\frac{1}{x}$ is not defined, in particular $\lim_{x\to 0^+}\frac{1}{x}=+\infty$, I have to take into account this fact or not? How can I treat $x=0$ in this case?
 A: The indefinite integral $\int f(x) \, dx$ is just a notation for the antiderivative, i.e. a function $F$ with $F' = f$. This implies that the domain of $F$ is a subset of the domain of $f$.
Your example function $f(x) = 1/x$ is defined on $\Bbb R\setminus \{ 0 \}$, therefore an antiderivative can only be defined on that set.
The notation $\int f(x) \, dx = F(x) + C$ is sometimes used to indicate that the “general antiderivative” is obtained by adding an arbitrary constant to $C$, e.g.
$$
\int \frac 1 x \, dx = \log |x| + C \, .
$$
As Hans Lundmark said, care must be taken if the domain of $f$ consists of multiple components. Strictly speaking, the general antiderivative of $1/x$ is
$$
 F(x) = \begin{cases}
  \log (-x) + C_1 & \text{ for } x < 0 \\
  \log (x) + C_2 & \text{ for } x > 0 \\
 \end{cases}
$$
with real constants $C_1, C_2$.
Back to your original question: If $f$ is not defined at a point $x_0$ then there is no antiderivative at $x_0$. But:

*

*If the domain of $f$ consists of multiple intervals then the constant of integration can be independently chosen on each interval on which the antiderivative exists.


*The fundamental theorem of calculus in the form $\int_a^b f(t) \, dt = F(b) - F(a)$ is only valid on each interval on which the antiderivative exists.
As an example, it would be wrong to say that $\int_{-1}^1 \frac 1x \, dx = \log|1| - \log |-1| = 0$.
