Boost Your Skills in Finding Antiderivatives

Antiderivative of unbounded function?
One way to visualize an antiderivative is that the area under the derivative is added to the initial value of the antiderivative to get the final value of the antiderivative over an interval.
The Riemann Series Theorem essentially says that you can basically get any value you want out of a conditionally convergent series by changing the order you add up the terms.
Now consider the function: $f(x)=1/x^3$
The function is unbounded over the interval $(-<!--\; -\; -->1,1)$, so it not integrable over this interval.
If you break f(x) into Riemann rectangles over the interval $(-<!--\; -\; -->1,1)$ and express the area as a Riemann sum, you essentially get a conditionally convergent series. And because of the Riemann Series Theorem, you can make the sum add up to anything. In other words, you can make the rectangles add up to whatever area you want by changing the order in which you add them up. This is in fact why a function needs to be bounded to be integrable - otherwise the area has infinite values/is undefined.
So my question is, in cases like this, how does the antiderivative "choose" an area? I mean in this case the antiderivative, $\frac{1}{-<!--\; -\; -->2x^2}$, choose to increase by $\frac{1}{-<!--\; -\; -->2(1{)}^2}-<!--\; -\; -->\frac{1}{-<!--\; -\; -->2(-<!--\; -\; -->1{)}^2}=0.$. In other words, the antiderivative "choose" to assign a area of zero to the area under $1/x^3$ from -1 to 1 even though the Riemann Series Theorem says the area can be assigned any value.
How did the antiderivative "choose" an area of 0 from the infinite possible values?

If a function is holomorphic in a certain domain, does it mean it necessarily has an antiderivative
Say I have a holomrphic function f(z) in domain D, say $|z|>4$ and my function is not holomorphic for $|z|\le <!--\; \le \; -->4$ - it has some poles there. Does it mean there is necessarily an antiderivative for f(z)?
I know I can use Morera's theorem and calculate all the residues inside $|z|\le <!--\; \le \; -->4$ and check if their sum is zero. However what seemed more intuitive to me is to prove that since ${\int <!--\; \int \; -->}_{\mathrm{\gamma <!--\; \gamma \; -->}}f(z)dz$ for every $\mathrm{\gamma <!--\; \gamma \; -->}$ contained in D, depends only on the start and end points, there exists an antiderivative for f(z).
Is my prof correct? Can I conclude that in general, a holomorphic function in a domain D not only has infinitely many derivatives in D, but it has also infinitely many antiderivatives there?

Must a Function be Bounded for the Antiderivative to Exist over a Given Interval?
In my Calculus class, we were given the following definition of antiderivative:
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 of f is denoted ${\displaystyle \int <!--\; \int \; -->f(x)dx}$.
My question is, don't we have to say that f should be bounded in the definition?
If not, then f is not integrable by definition, so we can't say anything about ${\displaystyle \int <!--\; \int \; -->f(x)dx}$, right?

Antiderivative of $\arctan <!--\; \; -->(-<!--\; -\; -->x^2)$.
As I said in the title I'm trying to find an antiderivative of $f(x)=\arctan <!--\; \; -->(-<!--\; -\; -->x^2)$.
I am aware that e.g. WolframAlpha can find one, but I have no clue how to do it by hand. Can anyone give me a hint?

How to find the general antiderivative of $f(x)=x(6-<!--\; -\; -->x{)}^2$?
I want to find the general antiderivative of $f(x)=x(6-<!--\; -\; -->x{)}^2$. However, I keep getting it wrong.
I am new to antiderivatives, but I think the first thing I should do is differentiate.
$\frac{d}{dx}(6-<!--\; -\; -->x{)}^2=2(6-<!--\; -\; -->x)\cdot <!--\; \cdot \; -->-<!--\; -\; -->1=-<!--\; -\; -->2(6-<!--\; -\; -->x)$
$f^{\prime }(x)=[(6-<!--\; -\; -->x{)}^2\cdot <!--\; \cdot \; -->1]+[-<!--\; -\; -->2(6-<!--\; -\; -->x)\cdot <!--\; \cdot \; -->x]=(6-<!--\; -\; -->x{)}^2-<!--\; -\; -->2x(6-<!--\; -\; -->x)$
According to the antiderivative power rule, when $n\ne <!--\; \ne \; -->-<!--\; -\; -->1$, $\int <!--\; \int \; -->x^{n}dx=\frac{x^{n+1}}{n+1}+C$.
So it seems like $\int <!--\; \int \; -->(6-<!--\; -\; -->x{)}^{2}dx=\frac{(6-<!--\; -\; -->x{)}^3}{3}+C$.
However, I can't find a rule that seems like it would work with $-<!--\; -\; -->2x(6-<!--\; -\; -->x)$. The closest thing that I can find is the "Multiplication by Constant Rule", $\int <!--\; \int \; -->cf(x)dx=c\cdot <!--\; \cdot \; -->\int <!--\; \int \; -->(f(x))dx$, but I'm not sure if I'm allowed to change $2x(6-<!--\; -\; -->x)$ into the form $2(6x-<!--\; -\; -->x^2)$.
$\int <!--\; \int \; -->2(6x-<!--\; -\; -->x^2)$
$=2\cdot <!--\; \cdot \; -->\int <!--\; \int \; -->(6x-<!--\; -\; -->x^2)$
$=2(6\int <!--\; \int \; -->x-<!--\; -\; -->\int <!--\; \int \; -->x^2)$
$=2(\frac{6x^2}{2}-<!--\; -\; -->\frac{x^3}{3})+C$
But $\frac{(6-<!--\; -\; -->x{)}^3}{3}-<!--\; -\; -->(6x^2-<!--\; -\; -->\frac{2x^3}{3})+C$ is incorrect.

Where is the absolute value when computing antiderivatives?
Here is a typical second-semester single-variable calculus question:
$\int <!--\; \int \; -->\frac{1}{\sqrt{1-<!--\; -\; -->x^2}}dx$
Students are probably taught to just memorize the result of this since the derivative of arcsin(x) is taught as a rule to memorize. However, if we were to actually try and find an antiderivative, we might let $x=\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}⟹<!--\; ⟹\; -->dx=\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}d\mathrm{\theta <!--\; \theta \; -->}$
so the integral may be rewritten as $\int <!--\; \int \; -->\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{\sqrt{1-<!--\; -\; -->{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}d\mathrm{\theta <!--\; \theta \; -->}=\int <!--\; \int \; -->\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{\sqrt{{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}d\mathrm{\theta <!--\; \theta \; -->}$
At this point, students then simplify the denominator to just $\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$, which boils the integral down to
$\int <!--\; \int \; -->1d\mathrm{\theta <!--\; \theta \; -->}=\mathrm{\theta <!--\; \theta \; -->}+C=\arcsin <!--\; \; -->x+C$
which is the correct antiderivative. However, by definition, $\sqrt{x^2}=|x|$, implying that the integral above should really be simplified to $\int <!--\; \int \; -->\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{|\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}|}d\mathrm{\theta <!--\; \theta \; -->}=\int <!--\; \int \; -->\pm <!--\; \pm \; -->1d\mathrm{\theta <!--\; \theta \; -->}$ depending on the interval for $\mathrm{\theta <!--\; \theta \; -->}$. At this point, it looks like the answer that we will eventually arrive at is different from what we know the correct answer to be.

Existence of antiderivative without Cauchy-Goursat's theorem
I wonder if anybody has tried the following kind of direct proof for the existence of an antiderivative of an analytic function on a star-shaped domain.
Theorem: Let $f:D\to <!--\; \to \; -->\mathbb{C}$ be an analytic function on a star-shaped domain D. Then f has an antiderivative F on D.
"proof": For simplicity, assume that every point in D is connected to $0\in <!--\; \in \; -->\mathbb{C}$ by a line segment. Define
$F(z)={\int <!--\; \int \; -->}_0^{1}zf(zt)dt.$
Then $\underset{h\to <!--\; \to \; -->0}{lim}\left(\frac{F(z+h)-<!--\; -\; -->F(z)}{h}\right)=\underset{h\to <!--\; \to \; -->0}{lim}{\int <!--\; \int \; -->}_0^1\left(\frac{(z+h)f(zt+ht)-<!--\; -\; -->zf(t)}{h}\right)dt.$
It can be checked that as $h\to <!--\; \to \; -->0$, the integrand converges to $\frac{d}{dt}tf(zt).$.
Therefore, $F^{\prime }(z)=f(z)$, provided that the limit and integral are interchangeable. qed.
Of course, a limit and integral cannot always be interchanged. But I wonder if anybody seriously considered the above line of proof.
Thanks. As far as I can search from several textbooks in complex variables, the above theorem is proved by using Cauchy-Goursat's theorem. More concretely, they use the equality ${\int <!--\; \int \; -->}_{z_0}^{z+h}f(z)dz-<!--\; -\; -->{\int <!--\; \int \; -->}_{z_0}^{z}f(z)dz={\int <!--\; \int \; -->}_z^{z+h}f(z)dz,$, which can be justified by Cauchy-Goursat's theorem. The point of my question is: Is it possible to directly invoke to the computation as above, without using Cauchy-Goursat's theorem? If this is possible, we get another proof of Cauchy-Goursat's theorem, at least for star-convex domains.

Jump discontinuity in antiderivative
I am dealing with integral $I={\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}dx\frac{{\sin }^2<!--\; \; -->x}{a^2+{\sin }^2<!--\; \; -->x}.$
I know the answer is $I=\mathrm{\pi <!--\; \pi \; -->}(1-<!--\; -\; -->\frac{a}{\sqrt{1+a^2}}).$.
However I am having some uncertainty getting there. I know that before plugging in the integration limits the solution looks like $I={[x-<!--\; -\; -->\frac{a{\tan }^{-<!--\; -\; -->1}<!--\; \; -->(\frac{\sqrt{1+a^2}}{a}\tan <!--\; \; -->x)}{\sqrt{1+a^2}}]}_{x=0}^{x=\mathrm{\pi <!--\; \pi \; -->}}$.
but at first sight this would yield merely $\mathrm{\pi <!--\; \pi \; -->}$. I can see that the right step to recover the correct solution is to take ${\tan }^{-<!--\; -\; -->1}<!--\; \; -->(\frac{\sqrt{1+a^2}}{a}\tan <!--\; \; -->x){|}_{x=\mathrm{\pi <!--\; \pi \; -->}}=\mathrm{\pi <!--\; \pi \; -->}$ however how do I justify this (besides that it yields the correct answer)?

Find an antiderivative of the function
Let $u:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a function. Find an antiderivative of $u^2{u}^{‴}$ in terms of u and its derivatives.
Note: I know an antiderivative of uu′′′ is $u{u}^{″}-<!--\; -\; -->\frac{u^{\prime 2}}{2}$. However it seems for this problem one needs to evaualate antiderivative of either uu′u′′ or $u^{\prime 3}$ which can't be computed easily via integration by parts

Operator on polynomials, antiderivative
We are given a linear map:
$\mathbb{R}[X]\ni <!--\; \ni \; -->p\to <!--\; \to \; -->q\in <!--\; \in \; -->\mathbb{R}[X],q^{\prime }=p,q(0)=0$ and two norms on $\mathbb{R}[X]$ : $||p|{|}_{\mathrm{\infty <!--\; \infty \; -->}}=\underset{t\in <!--\; \in \; -->[0,1]}{sup}|p(t)|$, $||p|{|}_1={\int <!--\; \int \; -->}_0^1|p(t)|dt$.
I want to check whether the map is continuous in these norms.
When it comes to the first norm, I get:
$||q|{|}_{\mathrm{\infty <!--\; \infty \; -->}}=\underset{t\in <!--\; \in \; -->[0,1]}{sup}|q(t)|$
By Lagrange mean value theorem we have that there exists $a\in <!--\; \in \; -->[0,1]$ such that $q^{\prime }(a)=\frac{q(t)-<!--\; -\; -->q(0)}{t}$, so $q(t)\le <!--\; \le \; -->\underset{[0,1]}{sup}|q^{\prime }(t)|$.
And $||p|{|}_{\mathrm{\infty <!--\; \infty \; -->}}=\underset{t\in <!--\; \in \; -->[0,1]}{sup}|p(t)|=\underset{t\in <!--\; \in \; -->[0,1]}{sup}|q^{\prime }(t)|\ge <!--\; \ge \; -->||q||$
Is that correct so far?

prove that ${\int <!--\; \int \; -->}_a^{b}f+{\int <!--\; \int \; -->}_{f(a)}^{f(b)}{f}^{-<!--\; -\; -->1}=bf(b)-<!--\; -\; -->af(a)$
Let $0<a<b$ let $f>0$ be continuous and strictly increasing function on [a,b]. Prove that ${\int <!--\; \int \; -->}_a^{b}f+{\int <!--\; \int \; -->}_{f(a)}^{f(b)}{f}^{-<!--\; -\; -->1}=bf(b)-<!--\; -\; -->af(a)$.
How to approach this problem . Any Hint? I am suppose to do it without using antiderivatives.

How does one go about evaluating the limits of nonelementary antiderivatives of functions?
We obviously know the following to be equal:
$\underset{z\to <!--\; \to \; -->\pm <!--\; \pm \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\text{erf}(z)=\pm <!--\; \pm \; -->1$
But as far as I'm aware, the only way we know how to calculate any given value of the error function is to approximate it. How do we know the limits at infinities without any way to evaluate it algabraically?

Antiderivative of $\frac{1}{4}\sin <!--\; \; -->(2x)$
I am trying to find the antiderivative of $\frac{1}{4}\sin <!--\; \; -->(2x)$ but i'mn ot sure how to find this. Can someone maybe give me what formula to use when it involves sin?

Antiderivative fraction
I'm would like to find the antiderivative of $\frac{x^2}{(x-<!--\; -\; -->1{)}^5}$
I tried without using partial fractions but I did not manage, so I started a lesson on partial fractions but I do not succeed in proving
$\frac{x²}{(x-<!--\; -\; -->1{)}^5}=\frac{1}{(x-<!--\; -\; -->1{)}^3}+\frac{2}{(x-<!--\; -\; -->1{)}^4}+\frac{1}{(x-<!--\; -\; -->1{)}^5}$
Could you tell me how to proceed?

Are these two antiderivatives equivalent?
Regarding the anti derivative of: ${\cos }^3<!--\; \; -->\left(x\right){\sin }^3<!--\; \; -->\left(x\right)$
Is the answer $-<!--\; -\; -->{\displaystyle \frac{{\sin }^4<!--\; \; -->\left(x\right)(2{\sin }^2<!--\; \; -->\left(x\right)-<!--\; -\; -->3)}{12}}$ equivalent to ${\displaystyle \frac{{\sin }^4<!--\; \; -->\left(x\right)}{4}}$.
If so, how? Integral Calculator says they're equivalent, but I don't understand how.

Show that two primitives/ antiderivatives are related via a constant
Let $I\subset <!--\; \subset \; -->R$ be an interval. A differentiable function $F:I\to <!--\; \to \; -->R$ is called a primitive for the function $f:I\to <!--\; \to \; -->\mathbb{R}$ if $F^{\prime }(x)=f(x)$ for all $x\in <!--\; \in \; -->I$.
Show: If $F_1$ and $F_2$ are two primitives for f on I then there is a constant $C\in <!--\; \in \; -->\mathbb{R}$ such that $F_2\equiv <!--\; \equiv \; -->F_1+C$, i.e. $F_2(x)=F_1(x)+C$ for all $x\in <!--\; \in \; -->I$.
What we have covered so far is the formal definition of the derivative in terms of the limit:
$\underset{h\to <!--\; \to \; -->0}{lim}\frac{F_1(x+h)-<!--\; -\; -->F_1(x)}{h}=f(x)$
$\underset{h\to <!--\; \to \; -->0}{lim}\frac{F_2(x+h)-<!--\; -\; -->F_2(x)}{h}=f(x)$
I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.
We normally just define $F_2=F_1+C$ and then it would follow immediately that these two functions have the same derivative.

Removing the $+c$ from the antiderivative
While I was first learning about integrals and antiderivatives, I took for granted that, for example, $\int <!--\; \int \; -->2xdx=x^2+c$. But if the constant can be any number, that means that $\int <!--\; \int \; -->2xdx=x^2=x^2+1$ which doesn't make any sense. I understand that the constant is usually just left as c, but it is implied that it can be any number, right? We don't define the square root of 4 to be 2 and -2 and we don't define arcsin(0) to be 0, $2\mathrm{\pi <!--\; \pi \; -->}$, $4\mathrm{\pi <!--\; \pi \; -->}$
If you try to define $\int <!--\; \int \; -->\frac{d}{dx}(f(x))dx=f(x)$, a problem arises. Mainly, if f(x) is something like $x^3+4$. The 4 gets lost when taking the derivative so we can't be sure of what a function is if we only know its derivative.
A potential solution to this problem would be to define a function S(f) such that it takes a function as its input and outputs a function stripped of its constant e.g. $S(x^3+4)=x^3$. If this is done, the antiderivative can be defined as follows:
$\int <!--\; \int \; -->\frac{d}{dx}(f(x))dx=S(f(x))$
$\frac{d}{dx}(\int <!--\; \int \; -->f(x)dx)=f(x)$
Is it possible to define S(f)? If not, is there any other way to define the antiderivative so that it doesn't have a constant?

Existence of an antiderivative in $U\cup V$
I am trying to prove the following:
Let $U,V\subset <!--\; \subset \; -->\mathbb{C}$ be open and connected sets such that $U\cap <!--\; \cap \; -->V$ is connected as well.
Let f be a holomorphic funtion with complex antiderivative on Uand V.
Then f has a complex antiderivative on $U\cup <!--\; \cup \; -->V$.
Let $F_1$ be the antiderivative on U and $F_2$ be the antiderivative on V then $F_1^{{}^{\prime }}=F_2^{{}^{\prime }}$ on the conncted set $U\cap <!--\; \cap \; -->V$.
Hence $F_1=F_2+c$.
I dont know how to get to get the antiderivative on $U\cup <!--\; \cup \; -->V$ from here.

Finding original functions of second derivatives expressions
I have an antiderivative problem.
Given some function:
$f^{″}(t)=t+\sqrt{t}$
And the conditions when the $t=1$:
$f(1)=1;$
$f^{\prime }(1)=2$
How do I go about deriving the original function given that I must do this with antiderivatives? My logic has been to just take the antiderivative of the original f′′(t) and then take the antiderivative of f′(t) but I am wrong in this approach?

Finding antiderivative problems
I'm having trouble understanding how to find these antiderivatives:
1. $e^2$
Is the antiderivative just $e^2\cdot <!--\; \cdot \; -->x+c$.
2. $f(x)=\sqrt[3]{x^2}+x\sqrt{x}$
$=x^{2^{\frac{1}{3}}}+x\cdot <!--\; \cdot \; -->x^{\frac{1}{2}}=x^{\frac{2}{3}}+x^{\frac{3}{2}}$
So the antiderivative according to antiderivative formulas:
$Anti=\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+\frac{x^{\frac{5}{2}}}{\frac{5}{2}}$
$=2\cdot <!--\; \cdot \; -->\frac{x^{\frac{5}{2}}}{\frac{5}{2}}$
$=4\cdot <!--\; \cdot \; -->\frac{x^{\frac{5}{2}}}{5}+C=F$
Is this right?
When I do a check, I don't get the original function:
$F^{\prime }=\frac{4}{5}\cdot <!--\; \cdot \; -->\frac{5}{2}\cdot <!--\; \cdot \; -->x^{\frac{3}{2}}=2\cdot <!--\; \cdot \; -->x^{\frac{3}{2}}$
But that doesn't equal the original function. What did I do wrong?