Boost Your Skills in Finding Antiderivatives

Textbooks that use notation with explicit argument variable in the upper bound ${\int <!--\; \int \; -->}^x$ for "indefinite integrals."
I dare to ask a question similar to a closed one but more precise.
Are there any established textbooks or other serious published work that use ${\int <!--\; \int \; -->}^x$ notation instead of $\int <!--\; \int \; -->$ for the so-called "indefinite integrals"?
(I believe I've seen it already somewhere, probably in the Internet, but I cannot find it now.)
So, I am looking for texts where the indefinite integral of cos would be written something like:
${\int <!--\; \int \; -->}^x\cos <!--\; \; -->(t)dt=\sin <!--\; \; -->(x)-<!--\; -\; -->C$
or ${\int <!--\; \int \; -->}^x\cos <!--\; \; -->(x)dx=\sin <!--\; \; -->(x)+C.$
(This notation looks more sensible and consistent with the one for definite integrals than the common one with bare $\int <!--\; \int \; -->$.)
IMO, the indefinite integral of f on a given interval I of definition of f should not be defined as the set of antiderivatives of f on I but as the set of all functions F of the form
$F(x)={\int <!--\; \int \; -->}_a^{x}f(t)dt+C,x\in <!--\; \in \; -->I,$,
with $a\in <!--\; \in \; -->I$ and C a constant (or as a certain indefinite particular function of such form). In other words, I think that indefinite integrals should be defined in terms of definite integrals and not in terms of antiderivatives. (After all, the integral sign historically stood for a sum.)
In this case, the fact that the indefinite integral of a continuous function f on an interval I coincides with the set of antiderivatives of f on I is the contents of the first and the second fundamental theorems of calculus:
1. The first fundamental theorem of calculus says that every representative of the indefinite integral of f on I is an antiderivative of f on I, and
2. The second fundamental theorem of calculus says that every antiderivative of f on I is a representative of the indefinite integral of f on I (it is an easy corollary of the first one together with the mean value theorem).

Antiderivatives and definite integrals - 2
Prove that the function $f(x)=x^2{\int <!--\; \int \; -->}_0^{1}t{\sin }^2<!--\; \; -->(tx)dt$
is differentiable in $\mathbb{R}$ and determine the formula f′(x) of its derivative.

Derivatives of Sets
So, I often see this:
$D_x[\int <!--\; \int \; -->f(x)]=f(x)$
But this is a derivative of a set of antiderivatives. What is the conceptual backing for this i.e. what does it mean to derive a set?

When ${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}|x{|}^{\mathrm{\alpha <!--\; \alpha \; -->}}|\tan <!--\; \; -->x{|}^{\mathrm{\beta <!--\; \beta \; -->}}dx<\mathrm{\infty <!--\; \infty \; -->}$ and ${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}|\tan <!--\; \; -->x{|}^{\mathrm{\gamma <!--\; \gamma \; -->}}dx<\mathrm{\infty <!--\; \infty \; -->}$? Closed forms?
For which $\mathrm{\alpha <!--\; \alpha \; -->}>0,\mathrm{\beta <!--\; \beta \; -->}\in <!--\; \in \; -->(0,2),\mathrm{\gamma <!--\; \gamma \; -->}\in <!--\; \in \; -->(0,2)$ do we have
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}|x{|}^{\mathrm{\alpha <!--\; \alpha \; -->}}|\tan <!--\; \; -->x{|}^{\mathrm{\beta <!--\; \beta \; -->}}dx<\mathrm{\infty <!--\; \infty \; -->}$
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}|\tan <!--\; \; -->x{|}^{\mathrm{\gamma <!--\; \gamma \; -->}}dx<\mathrm{\infty <!--\; \infty \; -->}?$
Can you also give a closed-form expression for the antiderivatives (or the definite integrals)?

Integratin $\int <!--\; \int \; -->{\tan }^3<!--\; \; -->(x)dx$ in two different ways gives two different answers
I was trying to find the antiderivative of a function $\int <!--\; \int \; -->{\tan }^3<!--\; \; -->(x)dx$.
However, due to substitution differences, my book has a answer of $\frac{1}{2}{\tan }^2<!--\; \; -->(x)+\ln <!--\; \; -->(\cos <!--\; \; -->x)+C$ while I got an answer $\frac{1}{2}{\sec }^2<!--\; \; -->(x)+\ln <!--\; \; -->(\cos <!--\; \; -->x)+C$.
The problem is what to substitute in $\int <!--\; \int \; -->\tan <!--\; \; -->x{\sec }^2<!--\; \; -->xdx$. The book puts $\tan <!--\; \; -->x=z$, while I put $\sec <!--\; \; -->x=z$. I don't know if both are correct. If they are, can all functions have multiple antiderivatives?

Distributional "Antiderivative"
Suppose that ${\mathbb{R}}^3$, fix $f\in <!--\; \in \; -->L^1({\mathbb{R}}^3;\mathbb{R})$ and let $g\in <!--\; \in \; -->L^1({\mathbb{R}}^3;{\mathbb{R}}^3)$ satisfies
$div(g)={\mathrm{\delta <!--\; \delta \; -->}}_a-<!--\; -\; -->f.$.
Then what is g? It what is the distributional anti-derivative of this thing?

Antiderivative of Antiderivative
Probably easy but I'm not very sure. If f(x) has an antiderivative F(x) then F(x) has also an antiderivative. True or False?

Antiderivative of $e^x/(1+2e^x)$
I know the solution is ${\displaystyle \frac{\ln <!--\; \; -->(2{\mathrm{e}}^x+1)}{2}}+C$.
However, the result I got was ${\displaystyle \frac{\ln <!--\; \; -->({\mathrm{e}}^x+0.5)}{2}}+C$.
What I did was:
$\begin{array}{rl}\int <!--\; \int \; -->\frac{e^x}{1+2e^x}dx& =\int <!--\; \int \; -->\frac{e^x}{2\ast <!--\; \ast \; -->(0.5+e^x)}dx\\ & =0.5\cdot <!--\; \cdot \; -->\int <!--\; \int \; -->\frac{e^x}{0.5+e^x}dx\\ & =0.5\cdot <!--\; \cdot \; -->(\ln <!--\; \; -->|0.5+e^x|+C).\end{array}$
I know there are antiderivative calculators online that show the correct method step by step, but I can't understand what I did wrong. What's the mistake?

Proof that two antiderivatives differ by a constant, quantifier confusion
The proof that two antiderivatives of f differ by a constant is given in several other questions.
By the mean value theorem, for all $a<b\in <!--\; \in \; -->I$, there exists an $x\in <!--\; \in \; -->[a,b]$ such that
$$\begin{gathered} (F-<!--\; -\; -->G{)}^{\mathrm{\prime <!--\; \prime \; -->}}(x)=\frac{(F-<!--\; -\; -->G)(b)-<!--\; -\; -->(F-<!--\; -\; -->G)(a)}{b-<!--\; -\; -->a} \\ =F^{\mathrm{\prime <!--\; \prime \; -->}}(x)-<!--\; -\; -->G^{\mathrm{\prime <!--\; \prime \; -->}}(x) \\ =f(x)-<!--\; -\; -->f(x) \\ =0 \\ ⟹<!--\; ⟹\; -->0=\frac{(F-<!--\; -\; -->G)(b)-<!--\; -\; -->(F-<!--\; -\; -->G)(a)}{b-<!--\; -\; -->a} \\ ⟹<!--\; ⟹\; -->(F-<!--\; -\; -->G)(b)=(F-<!--\; -\; -->G)(a) \\ ⟹<!--\; ⟹\; -->F(b)-<!--\; -\; -->G(b)=F(a)-<!--\; -\; -->G(a) \end{gathered}$$
so we can define
$C:=F(b)-<!--\; -\; -->G(b)=F(a)-<!--\; -\; -->G(a)$
But since that's true for all a and b, what happens if we let $a=infI$? Then $F(a)=G(a)=0,$, which means that $F(b)=G(b)$ for all $b>infI$. But that doesn't seem like it's true.
So I suspect I'm confused about quantifiers somewhere. Where am I going wrong?

Antiderivative of $e^{au}$
I cannot seem to figure out $\int <!--\; \int \; -->e^{au}du$. I have tried u-substitution (of course with a variable other than u) and can't get it to work out to the right answer. Any help is appreciated.

Do I have this idea of antiderivatives correct?
So is all it's saying that if there are two functions that have the same derivatives for every single x in the interval, then $f(x)=g(x)+\mathrm{\alpha <!--\; \alpha \; -->}$, means that the second function is just the exact same as f(x) except raised or lowered?

Smoothness properties of partial antiderivatives
I'm wondering what can be said about the well-behavior (or lack thereof) of antiderivatives of multivariable functions.
For example, consider the function $f(x,y)=2xy$. If we take the antiderivative of this function with respect to x, we obtain $F(x,y)=x^{2}y+h(y)$ where h(y) is an arbitrary function of y. The curious thing is there are no restrictions on what h could be. h could be a well-behaved differentiable function, or it could be a nasty function like the Weierstrass function, which is differentiable nowhere, or even the completely discontinuous rational number detector $h(x)=\{\begin{array}{ll}1& \text{if }x\text{ is rational}\\ 0& \text{if }x\text{ is irrational}\end{array}$
In either case, F would not be differentiable with respect to y, but even so, the partial derivative $\mathrm{\partial <!--\; \partial \; -->}F/\mathrm{\partial <!--\; \partial \; -->}x$ would be well-defined and will equal our original function f.
So obviously not every possible partial antiderivative will be differentiable with respect to the other variable, but clearly some are. If we choose $h(y)=y$, for example, then F is certainly differentiable with respect to y. In fact, as long as h is a differentiable function itself, F will be differentiable with respect to y.
My question is whether this holds true in general. If partial antiderivatives F exist for a multivariable function f with respect to one variable, is it always true that at least one such F will be differentiable with respect to the other variables?
I'm guessing the answer to this is No, but that raises the question of what requirements must we impose on the original function f to guarantee that a "nice" partial antiderivative exists. Is it enough that f be continuous? That f be fully differentiable? That f be continuously differentiable?
And even more generally, if I want to guarantee the existence of a partial antiderivative F that has some nice smoothness property (e.g. it is $C^2$, or $C^3$, etc.), what requirements must I impose on the original f to ensure this?

Evaluate $\int <!--\; \int \; -->\frac{1}{1+3{\sin }^2<!--\; \; -->x}dx$ (Making antiderivative continuous.)
Evaluate $\int <!--\; \int \; -->\frac{1}{1+3{\sin }^2<!--\; \; -->x}dx$
I know that this has an antiderivative on $\mathbb{R}$.
I can use the trig. substitution $t=\tan <!--\; \; -->x$ on $(-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+k\mathrm{\pi <!--\; \pi \; -->},\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+k\mathrm{\pi <!--\; \pi \; -->})$
$x=\arctan <!--\; \; -->t$
$dx=\frac{1}{1+t^2}dt$
To get $\int <!--\; \int \; -->\frac{1}{1+3{\sin }^2<!--\; \; -->x}dx$ = $\int <!--\; \int \; -->\frac{1}{1+3\frac{t^2}{t^2+1}}\cdot <!--\; \cdot \; -->\frac{1}{1+t^2}dt$= $\int <!--\; \int \; -->\frac{1}{1+(2t{)}^2}dt=\frac{1}{2}\int <!--\; \int \; -->\frac{1}{1+u^2}du=\frac{1}{2}\arctan <!--\; \; -->(2\tan <!--\; \; -->x)+C$
But this only appplies on the interval $(-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+k\mathrm{\pi <!--\; \pi \; -->},\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+k\mathrm{\pi <!--\; \pi \; -->})$, whereas the original function should have an antiderivative on $\mathbb{R}$. So I have to make it continuous, but I don't know how.
(I do know that we did a similar thing for $\int <!--\; \int \; -->|x|dx$, but I lost my notes and don't know how doing this is called in english so I can't google it. We called it "gluing" antiderivatives. And I remember that we utilized limits in some fashion related to the integration constants)

What does it mean to write an integral in its explicit form?
The first part of Fundamental Theorem of Calculus gives an abstract way of producing antiderivatives:
Write $F(x)={\int <!--\; \int \; -->}_1^x(t+1)dt$ in an explicit form. Check that its derivative is indeed $x+1$.
What does it mean to find an integral in it explicit form? Does it mean to find its anti derivative?

Antiderivative 1/z on $\mathbb{C}$
Let $\mathrm{\Omega <!--\; \Omega \; -->}\subseteq <!--\; \subseteq \; -->\mathbb{C}$ be open and $\mathrm{\gamma <!--\; \gamma \; -->}:[\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}]\to <!--\; \to \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ be a piecewise continuously differentiable and closed path.
Why does $z^{-<!--\; -\; -->1}$ have no antiderivative on $\mathbb{C}\setminus <!--\; \setminus \; -->0$?
Why is $\underset{\mathrm{\gamma <!--\; \gamma \; -->}}{\int <!--\; \int \; -->}{z}^{-<!--\; -\; -->1}dz=2\mathrm{\pi <!--\; \pi \; -->}i\cdot <!--\; \cdot \; -->{\text{ind}}_{\mathrm{\gamma <!--\; \gamma \; -->}}(0)$, where ${\text{ind}}_{\mathrm{\gamma <!--\; \gamma \; -->}}$ denotes the winding number.

Calculate limit involving the antiderivative of a function
Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$, $f(x)=e^{x^2}$. Now, consider F an antiderivative of f.
Calculate: $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{xF(x)}{f(x)}$.
I couldn't find the antiderivative of f, and besides finding F, I have no other ideas.

Show $(\genfrac{}{}{0ex}{}{n}{0})-<!--\; -\; -->\frac{1}{2}(\genfrac{}{}{0ex}{}{n}{1})+...+\frac{(-<!--\; -\; -->1{)}^n}{n+1}(\genfrac{}{}{0ex}{}{n}{n})=\frac{1}{n+1}$

Finding the Antiderivative of a complex function
I need to find the antiderivative of $f(z)=z\log <!--\; \; -->(z)$, but I am confused on how exactly to do that.
So we need to find $\int <!--\; \int \; -->z\log <!--\; \; -->(z)dz$ right? Since $z=x+iy$, then $\int <!--\; \int \; -->(x+iy)\log <!--\; \; -->(x+iy)(dx+idy)$, but then how do we split $\log <!--\; \; -->(x+iy)$?
Can we just integrate normally with respect to z THEN substitute with $z=x+iy$? We can use integration by parts as usual if so.

Confusion on ambiguity of antiderivatives.
Would this statement, $\mathrm{\forall <!--\; \forall \; -->}k\in <!--\; \in \; -->\mathbb{R}$, be true or false? And are these interchangeable under all circumstances?
$\int <!--\; \int \; -->f(x)dx+k=\int <!--\; \int \; -->f(x)dx$
Since we could assign F(x) to either the LHS or RHS, and differentiating F(x) would both result with f(x). However if we assign $I=\int <!--\; \int \; -->f(x)dx$, then $I=I+k\Rightarrow <!--\; \Rightarrow \; -->k=0$ even though we already stated "for all k". I think my misunderstanding stems from notation and what they represent. Would appreciate any clarification on this as I could not find a direct take on this flexibility of antiderivatives, especially in such context of equality.