Boost Your Skills in Finding Antiderivatives

Proving sin(1/x) has an antiderivative
I would want to prove that the function defined as follows: $f(x)=\sin <!--\; \; -->(1/x)$ and $f(0)=0$ has an antiderivative on the entire $\mathbb{R}$ (well, I'm not sure if I haven't worded this awkwardly, but essentially I am trying to show that f is a derivative of some function that is differentiable in every $x\in <!--\; \in \; -->\mathbb{R}$).
First off, f is continuous on $\mathbb{R}\setminus <!--\; \setminus \; -->0$, hence it has an antiderivative for every $x\in <!--\; \in \; -->\mathbb{R}\setminus <!--\; \setminus \; -->0$. However, f is not continuous in $x=0$. $x=0$
Furthermore, I'm curious about the significance of $f(0)=0$. Suppose we redefine our function, and take $f(0)=c$, for some $c\in <!--\; \in \; -->\mathbb{R}$, $c\ne <!--\; \ne \; -->0$. Would then f have an antiderivative on $\mathbb{R}$?

Why do the antiderivatives of $y=1$ follow $e^x$?
If you integrate $y=1$, you get x. If you integrate that, you get $\frac{x^2}{2}$, following the power rule. If you continue this, integrating over and over, the antiderivatives are: 1, x, $\frac{x^2}{2}$, $\frac{x^3}{6}$, $\frac{x^4}{24}$, and so on. It follows the pattern of $\frac{x^n}{n!}$, which happens to be the infinite polynomial of $e^x$. Just by integrating $y=1$, you follow the $exp()$ function. Is there any intuitive way why this is?

Antiderivative of $x^a{e}^{\left(bx\right)}$
I need to know the primitive function (Antiderivative) of this function:
$f(x)=x^a{e}^{\left(bx\right)}$ where a and b is a positive constants please how could I find the primitive of function ? is there any technique concerning this types?

Function with compact support whose iterated antiderivatives also have compact support
Notation: If $f:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is continuous, let us denote $If:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ its indefinite integral from 0, i.e., $(If)(x)={\int <!--\; \int \; -->}_0^{x}f(t)dt$, and iteratively $I^{k+1}f=I(I^{k}f)$.
Remark: If f is a continuous function with support contained in the open interval ]0,1[ then If has support contained in ]0,1[ iff $(If)(1)=0$.
Main question: Does there exist a $C^{\mathrm{\infty <!--\; \infty \; -->}}$ function f with support contained in the open interval ]0,1[ such that $I^{k}f$ has support contained in ]0,1[ for every $k\ge <!--\; \ge \; -->0$, or, equivalently, $(I^{k}f)(1)=0$ for all $k\ge <!--\; \ge \; -->0$?
Equivalent formulation: Does there exists a sequence $(f_k{)}_{k\in <!--\; \in \; -->\mathbb{Z}}$ of $C^{\mathrm{\infty <!--\; \infty \; -->}}$ functions each with support contained in the open interval ]0,1[, such that $f_{k-<!--\; -\; -->1}$ is the derivative of $f_k$?
Weaker question: Does there at least exist a continuous function f with the properties demanded in the main question?
Stronger question: Does there exist a $C^{\mathrm{\infty <!--\; \infty \; -->}}$ function f with compact support, whose Fourier transform vanishes identically on a nontrivial interval?
(A positive answer to the latter would imply a positive answer to the main question: rescale the function so its support is contained in ]0,1[, multiply it appropriately so its Fourier transform vanishes in a neighborhood of 0, and observe that the Fourier transform of $I^{k}f$ is, up to constants, ${\mathrm{\xi <!--\; \xi \; -->}}^k$ times that of f.)
Edit: Before someone points out that the identically zero function fits the bill, I should add that I want my functions to not vanish identically.

Does equality of antiderivatives imply equality almost everywhere?
If two Lebesgue integrable functions $f,g:[a,b]\to <!--\; \to \; -->\mathbb{R}$ satisfy
${\int <!--\; \int \; -->}_a^{x}f(s)ds={\int <!--\; \int \; -->}_a^{x}g(s)ds,$
for every $x\in <!--\; \in \; -->[a,b],$, then is is it true that $f(x)=g(x)$ almost everywhere?

Question about constants appearing in antiderivatives
Say we have two functions f(x) and g(x) where $\frac{df(x)}{dx}=\frac{dg(x)}{dx}$. This surely means that we can write $f(x)=g(x)+C$. Now let's say we are graphing these two functions and we want to edit one of them, say f(x) in such a way that the graph of f(x) lies perfectly on the graph of g(x). This would be possible to do by adding C to f(x) however only if C is a rational constant; if it is an irrational constant then we would need to use infinitely many digits of C to accomplish our goal since if we are using a finite number of digits of C it would be possible to "zoom in" the graph and see that the two functions don't overlap. This is only the context from which the question I want to ask emerged.
What makes this constant irrational? Is the rationality/irrationality of the constant somehow dependent on the nature of the derivative of f(x) and g(x)?

Showing that some Holomorphic Functions have Antiderivatives
Show that there is a holomorphic function defined in the set
$\mathrm{\Omega <!--\; \Omega \; -->}=\{z\in <!--\; \in \; -->\mathbb{C}:|z|>4\}$
whose derivative is $\frac{z}{(z-<!--\; -\; -->1)(z-<!--\; -\; -->2)(z-<!--\; -\; -->3)}$
Is there a holomorphic function on $\mathrm{\Omega <!--\; \Omega \; -->}$ whose derivative is
$\frac{z^2}{(z-<!--\; -\; -->1)(z-<!--\; -\; -->2)(z-<!--\; -\; -->3)}?$
I know that each holomorphic function on an open convex set has an antiderivative, but $\mathrm{\Omega <!--\; \Omega \; -->}$ isn't convex. Should I look for functions with these derivatives, such as $f(z)=\frac{1}{2}(\log <!--\; \; -->(1-<!--\; -\; -->z)-<!--\; -\; -->4\log <!--\; \; -->(2-<!--\; -\; -->z)+3\log <!--\; \; -->(3-<!--\; -\; -->z))$ and $g(z)=\frac{1}{2}(9\log <!--\; \; -->(z-<!--\; -\; -->3)-<!--\; -\; -->8\log <!--\; \; -->(z-<!--\; -\; -->2)+\log <!--\; \; -->(z-<!--\; -\; -->1))$, or are there any well-known propositions on finding antiderivatives in open, non-convex sets?

Why isn't there a good product formula for antiderivatives?
Computing antiderivatives is more challenging than computing derivatives, in part due to the lack of a "product formula''; namely, while (fg)′ can be expressed in terms of f, f′, g, g′, there seems to be no way to express $\int <!--\; \int \; -->fg$ in terms of $f,\int <!--\; \int \; -->f,g\int <!--\; \int \; -->g$ and related quantities. Is there any intuitive reason or heuristic explanation for why no such formula exists? I'm looking for a non-rigorous explanation which can be understood by first-year calculus students.

Definition of antiderivative - is it okay for f to be non-integrable?
I'm trying to figure out if the following definition of antiderivative is correct:
Let f be a function defined on an interval. An antiderivative of f is any function F such that $F^{\prime }=f$. The collection of all antiderivatives is denoted ${\displaystyle \int <!--\; \int \; -->f(x)dx}$.
Is it okay to have no conditions on f that ensure it's integrable?

Antiderivative Theory Problem
A function f is differentiable over its domain and has the following properties:
1. ${\displaystyle f(x+y)=\frac{f(x)+f(y)}{1-<!--\; -\; -->f(x)f(y)}}$.
2. $\underset{h\to <!--\; \to \; -->0}{lim}f(h)=0$
3. $\underset{h\to <!--\; \to \; -->0}{lim}f(h)/h=1$.
i) Show that $f(0)=0$.
ii) show that $f^{\prime }(x)=1+[f(x){]}^2$ by using the def of derivatives Show how the above properties are involved.
iii) find f(x) by finding the antiderivative. Use the boundary condition from part (i).
So basically I think I found out how to do part 1 because if $x+y=0$ then the top part of the fraction will always have to be zero.
part 2 and 3 are giving me trouble. The definition is the limit $(f(x+h)-<!--\; -\; -->f(x))/h$.
So I can set $x+y=h$ and make the numerator equal to f(h)?

Antiderivative of discontinuous function
I am having confusion regarding anti-derivative of a function.
$f(x)=\{\begin{array}{ll}-<!--\; -\; -->\frac{x^2}{2}+4& x\le <!--\; \le \; -->0\\ \phantom{-<!--\; -\; -->}\frac{x^2}{2}+2& x>0\end{array}$
Consider the domain [-1,2]. Clearly the function is Riemann integrable as it is discontinuous at finite number of point. However is there a function g(x) such that $g^{\prime }(x)=f(x)\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->[-<!--\; -\; -->1,2]$?

Are both of these indefinite integrals?
Scenario 1: $\begin{array}{}\text{(1)}& f(x):=\int <!--\; \int \; -->\cos <!--\; \; -->(x)dx\end{array}$
Scenario 2: $$\begin{gathered} \begin{array}{}\text{(2)}& g(x):={\int <!--\; \int \; -->}_a^x\cos <!--\; \; -->(t)dt \\ [\text{edited after seeing @hardmath's comment}] \\ [\text{a is a constant}]\end{array} \end{gathered}$$
Comments:
Both (1) and (2) are functions. An antiderivative/indefinite integral is a function, while a definite integral is a number. So, $$\begin{gathered} \begin{array}{}\text{(3)}& f(x)=g(x)=\sin <!--\; \; -->(x)+c \\ [\text{c is a constant}]\end{array} \end{gathered}$$
Is line (3) correct? Are both (1) and (2) antiderivatives/indefinite integrals?

Proof: Antiderivative of a function differ by a constant
Let f(x) a function defined at $I\subseteq <!--\; \subseteq \; -->\mathbb{R}$ and assume that F(x) and G(x) are the antiderivatives of f(x) in I, so there is a c such that for all $x\in <!--\; \in \; -->I$, $F(x)=G(x)+c$.
Let us define $H(X)=F(x)-<!--\; -\; -->G(X)$ therefore $H^{\prime }(X)=F^{\prime }(x)-<!--\; -\; -->G^{\prime }(X)=f(x)-<!--\; -\; -->f(x)=0$. Which theorem should be used to finalize the proof?

Composing a function with its antiderivative
What does it mean that, given a function and its antiderivative, if I make the composition $g(h(x))=h(g(x))=\mathrm{\alpha <!--\; \alpha \; -->}(x)$? I mean, I was thinking that $\mathrm{\alpha <!--\; \alpha \; -->}(x)$ could be some sort of identity function.
For example: Given, $g(x)={\displaystyle \frac{x^3}{3}}$, $h(x)=x^2$, then $g(h(x))={\displaystyle \frac{x^6}{3}}=h(g(x))$ where ${\displaystyle \frac{x^6}{3}}=\mathrm{\alpha <!--\; \alpha \; -->}(x)$

Finding the antiderivative
Let $\int <!--\; \int \; -->\frac{1+\frac{1}{x^2}}{x^2-<!--\; -\; -->1+\frac{1}{x^2}}dx$
I've been told use the substitution: $t=\frac{x-<!--\; -\; -->1}{x}$.
But how to apply it on the integral?

Indefinite integral $\int <!--\; \int \; -->\frac{1}{2-<!--\; -\; -->\cos <!--\; \; -->(x)}dx$ has discontinuities. How to fix?
Using the standard tangent half-angle substitution, we get
$\int <!--\; \int \; -->\frac{1}{2-<!--\; -\; -->\cos <!--\; \; -->(x)}dx=\frac{2}{\sqrt{3}}{\tan }^{-<!--\; -\; -->1}<!--\; \; -->(\sqrt{3}\tan <!--\; \; -->\frac{x}{2})+C$
The resulting antiderivatives are piecewise continuous functions with discontinuities at $x=(2k+1)\mathrm{\pi <!--\; \pi \; -->}$. However, the integrand is continuous everywhere, so any antiderivative must be continuous everywhere (??) according to FTC. So, how do we deal with this problem? What is the correct form of the antiderivatives? Is there a form that avoids the discontinuities?

Group of polynomials with real coefficients under adition - antiderivative that passes through (0,0)?
Let G be the group of all polynomials with real coefficients under addition. For each f in G, let ∫f denote the antiderivative of f that passes through the point (0,0). Show that the mapping $f\to <!--\; \to \; -->\int <!--\; \int \; -->f$ from G to G is a homomorphism. What is the kernel of this mapping? Is this mapping a homomorphism if ∫f denotes the antiderivative of f that passes through (0,1)?
My question here is what does the bolded sentence mean? Does it mean the n-th antiderivative of f such that $(x,y)=(0,0)$ is a solution after n consecutive antiderivatives?

A function not continuous but sill has antiderivative (necessary and sufficient condition for having antiderivative)
We already know:
1) if f(x) continuous on domain D then it is integrable and has antiderivative.
2) f(x) is almost continuous (that is, the set of discontinuities has measure zero) is equivalent to integrability, but "almost continuous" doesn't guarantee having antiderivative.
Now I am wondering is it possible for f(x) not continuous but still have antiderivative. Could you give me an example. It seems that people don't care about a function that have antiderivative, just integrable functions. Thank you.

Is there a formal definition for antiderivatives?
In the way the derivative can be defined as a limit, specifically
$f^{\prime }(x):=\underset{h\to <!--\; \to \; -->0}{lim}\frac{f(x+h)-<!--\; -\; -->f(x)}{h}$ or any of the other possible variants, is there a way to define the antiderivative, as in indefinite integral?
The handful of sources I've looked over (Wikipedia and MathWorld, to name a few) all refer to the antiderivative simply as a "nonunique inverse operator" (I'm paraphrasing). I can't say I'm completely satisfied with this notion. Is the following the best we can do?
If F(x) is a function that satisfies ${\displaystyle \frac{d}{dx}}F(x)=f(x)$, then F(x) is called an antiderivative of f(x). (Forgive the lack of formality.)

Antiderivative of piecewise function
I want to calculate the anti-derivative of:
$f(x)=\{\begin{array}{ll}-<!--\; -\; -->\sin <!--\; \; -->(x)& \text{if }x\ge <!--\; \ge \; -->0\\ 1-<!--\; -\; -->x^2& \text{if }x<0\end{array}$
such as $F(\mathrm{\pi <!--\; \pi \; -->}/2)=0$
I calculated both antiderivatives, thus I did get cos(x) for the first one and $x-<!--\; -\; -->x^3/3$. Question is to calculate $F(\mathrm{\pi <!--\; \pi \; -->})$ + $F(-<!--\; -\; -->1)$. But If I calculate that with null constant, I get $-<!--\; -\; -->5/3$ which is wrong.
Then could you help me on which value should I give to the constants $C_1+C_2$?