Boost Your Skills in Finding Antiderivatives

Antiderivatives such as $\int <!--\; \int \; -->\frac{dx}{\sqrt{e^{2x}+c}}$

Antiderivative of lebesgue integrable function
Let $f:[0,b]\to <!--\; \to \; -->\mathbb{R}$ be Lebesgue integrable. We define
$g(x)={\int <!--\; \int \; -->}_x^b\frac{f(t)}{t}dt,0<x\le <!--\; \le \; -->b.$
Show that g(x) is Lebesgue integrable in [0,b]. Also show that
${\int <!--\; \int \; -->}_0^{b}g(x)dx={\int <!--\; \int \; -->}_0^{b}f(t)dt.$
If we show the asked equality then it is obvious that g is Lesbegue integrable, since f is. Now, I have try to show the equality using the fact that,
${\int <!--\; \int \; -->}_0^b[g^{\prime }(x)-<!--\; -\; -->\frac{f(x)}{x}]dx=0$, since g is the antiderivative of f but I had no success. I am pleased to know other ideas to approach this problem.

F is integrable but has no indefinite integral
Let $f(x)=\{\begin{array}{ll}0,& x\ne <!--\; \ne \; -->0\\ 1,& x=0.\end{array}$
Then f is clearly integrable, yet has no antiderivative, on any interval containing 0, since any such antiderivative would have a constant value on each side of 0 and have slope 1 at 0 an impossibility.
So does this mean that f has no indefinite integral?
EDIT
My understanding is that the indefinite integral of f is the family of all the antiderivatives of f, and conceptually requires some antiderivative to be defined on the entire domain. Is this correct?

Is there a way to find "polynomial rational functions" having point of inflections, in their graphs?
For example if the derivative of f(x) is
$f^{\prime }(x)=\frac{(x-<!--\; -\; -->1{)}^2(x-<!--\; -\; -->3)}{(x-<!--\; -\; -->2)}$
then f(x) has a point of inflection at $x=1$.
But f(x) is not a polynomial rational !
Is there a way to determine which antiderivatives are the ratio of to polynomials?

Looking for antiderivative
Find $\int <!--\; \int \; -->\log <!--\; \; -->(\sqrt{x+1}+\sqrt{x})dx$
I used the subtitution $\sqrt{x}=\tan <!--\; \; -->y$.
but i the integral became too long and involves $\int <!--\; \int \; -->{\sec }^3<!--\; \; -->ydy$
and also $\int <!--\; \int \; -->{\sec }^5<!--\; \; -->ydy$
but it solvable so i search for a simple solution if exist? Note log is the natural logarithm

Two apparently different antiderivatives of $\frac{1}{2x}$
What is right way to calculate this integral and why?
$\int <!--\; \int \; -->\frac{1}{2x}\text{d}x$
I thought, that this substitution is right:
$t=2x$
$\text{d}t=2\text{d}x$
$\frac{\text{d}t}{2}=\text{d}x$
$\int <!--\; \int \; -->\frac{1}{2x}\text{d}x=\int <!--\; \int \; -->\frac{1}{t}\frac{\text{d}t}{2}=\frac{1}{2}\ln <!--\; \; -->|2x|+C.$
But it's not right, because this is the correct answer:
$\int <!--\; \int \; -->\frac{1}{2x}\text{d}x=\frac{1}{2}\int <!--\; \int \; -->\frac{1}{x}\text{d}x=\frac{1}{2}\ln <!--\; \; -->|x|+C.$
Can someone explain me, why is the first way wrong?

Definition of setting the integration constant equal to zero.
In Calculus books, we often hear of setting the integration constant equal to zero. But what does that actually mean? For example, the set of antiderivatives of 2x is $x^2+C$. If we set the constant equal to zero, we get simply $x^2$. However, I can also write the set of antiderivatives as $x^2+\mathrm{\pi <!--\; \pi \; -->}+C$. In that case, setting the constant equal to zero would give us $x^2+\mathrm{\pi <!--\; \pi \; -->}$. So, my question is, what does setting the constant equal to zero in an antiderivative actually mean?

How to prove this function has Darboux's property?
$f(x)=\{\begin{array}{ll}\cos <!--\; \; -->(\frac{1}{x})& \text{if }x\ne <!--\; \ne \; -->0\\ 0& \text{if }x=0\end{array}$
How do I prove this function has Darboux's property? I know it has it because it has antiderivatives, but how do I prove it otherwise, with intervals maybe?

Can we allow f to be undefined at finitely many points in (a,b) when formulating ${\int <!--\; \int \; -->}_a^{b}f(x)dx=F(b)-<!--\; -\; -->F(a)$, (F is the antiderivative of f)
Let f be a real-valued function on a closed interval [a,b] undefined only at finite points in (a,b). Let F be antiderivative of f. Then:
${\int <!--\; \int \; -->}_a^{b}f(x)dx=F(b)-<!--\; -\; -->F(a)$
Is the theorem true? How shall we prove it?
I ask this because I have read that function undefined at finite points in the interval (a,b) doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral.

Antiderivative of $|\sin <!--\; \; -->(x)|$
Because ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\sin <!--\; \; -->(x)\mathrm{d}x=2,$, then ${\int <!--\; \int \; -->}_0^{16\mathrm{\pi <!--\; \pi \; -->}}|\sin <!--\; \; -->(x)|\mathrm{d}x=32.$.
And Wolfram Alpha agrees to this, but when I ask for the indefinite integral $\int <!--\; \int \; -->|\sin <!--\; \; -->(x)|\mathrm{d}x,$,
Wolfram gives me $-<!--\; -\; -->\cos <!--\; \; -->(x)\mathrm{s}\mathrm{g}\mathrm{n}(\sin <!--\; \; -->(x))+c.$.
However, $[-<!--\; -\; -->\cos <!--\; \; -->(x)\mathrm{s}\mathrm{g}\mathrm{n}(\sin <!--\; \; -->(x)){]}_0^{16\mathrm{\pi <!--\; \pi \; -->}}=0,$,
So what's going on here? What is the antiderivative of $|\sin <!--\; \; -->(x)|$?

Antiderivative of a polynomial fraction
I am trying to solve this problem and I can't seem to understand where I am going wrong. My understanding of antiderivatives is this formula:
$\int <!--\; \int \; -->x^n\mathrm{d}x=\frac{x^{n+1}}{n+1}+c$
The problem I am trying to solve is:
$f(x)=\frac{x^5-<!--\; -\; -->x^3+8x}{x^4}$
I think there are two possible parts that I am getting wrong, I started the problem by separating them into individual fractions and solving each fraction at a time. I think I get it wrong at either the ln(x) or $\frac{8x^{-<!--\; -\; -->2}}{-<!--\; -\; -->2}$. Here is my attempt: $f(x)=\frac{x^5}{x^4}-<!--\; -\; -->\frac{x^3}{x^4}+\frac{8x}{x^4}=x-<!--\; -\; -->\frac{1}{x}+\frac{8}{x^3}$
Then from here I get the antiderivative:
$F(x)=\frac{x^2}{2}-<!--\; -\; -->\ln <!--\; \; -->(x)+\frac{8x^{-<!--\; -\; -->2}}{-<!--\; -\; -->2}+c=\frac{x^2}{2}-<!--\; -\; -->\ln <!--\; \; -->(x)-<!--\; -\; -->\frac{4}{x^2}+c$
I keep getting the wrong answer when I plug it in, am I missing some fundamental piece of information about antiderivatives which is causing me to mess up the ln(x) or is my algebra wrong and I mess up around $\frac{8x^{-<!--\; -\; -->2}}{-<!--\; -\; -->2}$?

Complex antiderivative
I am confused on a couple things:
1.) Why is it that an integral of a complex valued function of a complex variable exists if f(z(t)) is piecewise continuous (and/or piecewise continuous on $\mathbb{C}$) and not continuous, like a real?
2.) Why is it that one cannot make use of an antiderivative to evaluate an integral of a function like 1/z, on a contour of something like $z=2e^{i\mathrm{\theta <!--\; \theta \; -->}}$ positively oriented with $-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->\mathrm{\pi <!--\; \pi \; -->}$? That is, because $F^{\prime }(z)=1/z$ is undefined at 0, it is disqualified (I think). But why?

Show that one-form got an antiderivative
Let $F:{\mathbb{R}}^d\setminus <!--\; \setminus \; -->\{0\}\to <!--\; \to \; -->{\mathbb{R}}^d$ be a continous vector field with
$F(x)=\mathrm{\phi <!--\; \phi \; -->}(\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}^2)x$ for a continous $\mathrm{\phi <!--\; \phi \; -->}:(0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$. Show that the one-form ${\mathrm{\omega <!--\; \omega \; -->}}^F(p)(v):=⟨<!--\; ⟨\; -->F(p),v⟩<!--\; ⟩\; -->$, got an antiderivative.
I have no idea. Do i have to show that ${\mathbb{R}}^d\setminus <!--\; \setminus \; -->\{0\}$ is star-like and that F is closed?

Conventions for antiderivatives notation.
Am I setting myself up for a fundamental misconception if I consider antidifferentiation denoted by $\int <!--\; \int \; -->$ sign as the operation that is the inverse of the differential of a function denoted by d and that also returns an arbitrary constant C? To make things clear suppose I have a function F where $\frac{d}{dx}F(x)=f(x)$ for all x in some interval. Then it follows $dF(x)=f(x)dx$ now if I have $\int <!--\; \int \; -->(dF(x))=F(x)+C$ from my definition of antiderivative operation denoted by $\int <!--\; \int \; -->$ then it also means $\int <!--\; \int \; -->f(x)dx=F(x)+C$. What do you guys think?

Antiderivative of log(x) without Parts
I understand how the antiderivative of log(x) can be obtained by Integration by Parts (i.e. product rule), but I was wondering how-if at all- it could be obtained only using sum/difference rule and substitution/chain rule.

Determine antiderivative of f.
So I have a function:
$f(t)=\{\begin{array}{ll}-<!--\; -\; -->1,& \text{if }-<!--\; -\; -->2\le <!--\; \le \; -->t\le <!--\; \le \; -->0\\ 2-<!--\; -\; -->\sqrt{t^2-<!--\; -\; -->6t+9},& \text{if }0<t<6\\ -<!--\; -\; -->1,& \text{if }6\le <!--\; \le \; -->t\le <!--\; \le \; -->8\end{array}$
I need to find the antiderivative of f(t) on the interval: $-<!--\; -\; -->2<t<8$.
And I need to investigate whether the antiderivative has a maximum value on the interval.
So far I've figured out that the second function $2-<!--\; -\; -->\sqrt{t^2-<!--\; -\; -->6t+9}$ can be described as an absolute value: $2-<!--\; -\; -->|t-<!--\; -\; -->3|$.
That makes it 4 different functions in total that describes f(t)
However, I'm not sure exactly how to formulate a function for the antiderivative on the given interval.

Antiderivative of $8t^{-<!--\; -\; -->1/2}$
I am trying to find the antiderivative of $8t^{-<!--\; -\; -->1/2}$ but im just not understanding how to do this.
I saw someone get out of $t^{1/2}$ the answer $2t^{1/2}$. Can someone help me out?

Find an antiderivative for $\sqrt{z}$
I'm having some trouble on this homework problem:
If $\sqrt{z}$ is defined by $\sqrt{z}=e^{(\log <!--\; \; -->z)/2}$ for the branch of the log function defined by the condition $-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}/2\le <!--\; \le \; -->\arg <!--\; \; -->(z)\le <!--\; \le \; -->3\mathrm{\pi <!--\; \pi \; -->}/2$, find an antiderivative for $\sqrt{z}$ and then find ${\int <!--\; \int \; -->}_{\mathrm{\gamma <!--\; \gamma \; -->}}\sqrt{z}dz$, where $\mathrm{\gamma <!--\; \gamma \; -->}$ is any path from -1 to 1 which lies in the upper half-plane.
I think the second part will be relatively straightforward by FTC once I find an antiderivative. What I'm having trouble with is the first part. My intuition from real-valued calculus says that an antiderivative for $\sqrt{z}$ is $2/3z^{3/2}$. I'm not sure how to show this in the complex setting though. How do I use the log function and the associated branch to show this? Thanks.

How to reconcile the fact that the antiderivative of sin(x) cos(x) has two possible answers?
So here is the problem.
$$\begin{gathered} \int <!--\; \int \; -->\sin <!--\; \; -->(x)\cos <!--\; \; -->(x)dx=\frac{1}{2}{\sin }^2<!--\; \; -->(x)+c_1 \\ =-<!--\; -\; -->\frac{1}{2}{\cos }^2<!--\; \; -->(x)+c_2 \end{gathered}$$
This fact doesn't make much sense to me. How do you reconcile the fact that there are two possibilities (notice that these answers aren't only different by a constant)?
What are some other functions that have more than one antiderivatives (not only different by a constant)?

Is there a general formula for antiderivatives the same way there is for derivatives?
So, I know that first-order derivatives can be calculated with the formula:
$f^{\prime }(x)=\underset{h\to <!--\; \to \; -->0}{lim}\frac{f(x+h)-<!--\; -\; -->f(x)}{h}$
Is there such a limit/standard form for antiderivatives?