If a function is bounded on a closed interval, but there are infinite discontinuities, is the function necessarily not integrable? If it is integrable, what conditions should it satisfy

If a function is bounded on a closed interval, but there are infinite discontinuities, is the function necessarily not integrable? If it is integrable, what conditions should it satisfy

If there is only one accumulation point for these infinite discontinuities, then the function is integrable

The elementary function f (x) in its defined interval [a, b] may not be () a continuous B differentiable C, there exists the original function d integrable

The most obvious example is the elementary function y = √ x ^ 2 = | x |, whose domain is r, but it is not differentiable at x = 0

Bounded function and unbounded function (two questions)
Let a set M and an interval x = [a, b]
Function y = f (x)
A more popular understanding:
Bounded function: no matter x takes any number in the interval of X, the value of Y is in the range of M, then f (x) is said to be bounded in X, and vice versa
But when I saw two questions, I was confused as follows
1. In the interval (0, + (horizontal 8)), the unbounded of the following functions is () (+ (horizontal 8) denotes positive infinity)
A.y = e^-x^2 B.y=1/(1+x^2)
C.cos x D.y = x sin x
Where is m?
2. In the following interval, f (x) = LG (x + 1) is bounded ()
A.(-1,2) B.(-1,10^100)
C. (0,3) d. (0, + (inverted 8))
I don't know where m is

"Bounded function: no matter x takes any number in the interval of X, the value of Y is in the range of M, then f (x) is said to be bounded and unbounded in X, and vice versa." wrong definition, it should be the upper bound of set: for subset B of ordered set a (which may be totally ordered set or partially ordered set), if for any B ∈ B, there exists M