Master Calculus 1 Practice Problems with Expert Help

Continued Fractions Approximation
$\frac{x^2+3x+2}{x^2-<!--\; -\; -->x+1}$

How is the integral calculator finding this ${\int <!--\; \int \; -->}_0^a\frac{a^2-<!--\; -\; -->x^2}{(a^2+x^2{)}^2}dx$ while I get an undefined expression?

I want to evaluate the following integral:
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^0\frac{R}{{\sqrt{x^2+R^2}}^3}\mathrm{d}x$

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?

How do you find the instantaneous slope of y=4 at x=100?

If mn$\frac{m}{n}$ is an approximation to $\sqrt{2}$ then prove that $\frac{m}{2n}+\frac{n}{m}$ is a better approximation to $\sqrt{2}.$.(where mn is a rational number)

I am trying to calculate the integral
$\int <!--\; \int \; -->\frac{dx}{\sqrt[3]{(x+1{)}^2(x-<!--\; -\; -->1{)}^7}}.$

How do you find the instantaneous rate of change of the function $y=4x^3+2x-<!--\; -\; -->3$ when x=2?

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.

How to find instantaneous rate of change for $s(t)=t^3+t$ when t=2?

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 find instantaneous rate of change for $x^3+2x^2+x$ at x from 1 to 2?

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?

Evaluate $\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{e^{n^2}}{(2n)!}$

Lower estimate for ${\int <!--\; \int \; -->}_0^{T}f(x)\sin <!--\; \; -->f(x)dx$

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?