Master Calculus 1 Practice Problems with Expert Help

How does one go from this: $\int <!--\; \int \; -->\frac{l_0(1-\xi /x_1)}{\sqrt{{\kappa }_0^2(1-x_0/x_1{)}^2-l_0^2(1-\xi /x_1{)}^2}}d\xi$

Implication of complex-differentiability on the existence of antiderivatives
This is something that came to me randomly:
We know that a complex-differentiable function is infinitely-differentiable, and that elementary complex functions follow the same formal rules for differentiation as elementary real functions.
Does this mean that whenever we have a function defined via elementary functions on R whose derivative is not differentiable everywhere on R, we can conclude that this function does not have an antiderivative that can be expressed in terms of elementary functions?
For example: the function $x^{3}sin(\frac{1}{x^2})$ extended by zero at the origin. This function is differentiable everywhere on R, but its derivative is not continuous (thus not differentiable) at the origin.
Thus if $x^{3}sin(\frac{1}{x^2})$ had an antiderivative expressible in terms of elementary functions, we would obtain a contradiction with the fact that differentiable complex functions are smooth.
Here are my questions:
1) Is my reasoning sound?
2) Is this discussed somewhere in more generality, and has it been formalized in some way?
3) Does anyone have any examples where this observation leads to interesting conclusions?

Help me understand this antiderivative
I have a simple equation I typed into WolframAlpha to see what it was. Look what WO gave me:
Indefinite integrals: $\int <!--\; \int \; -->\frac{1}{2\sqrt{u}}du=\sqrt{u}+\text{constant}.$
Possible intermediate steps:
$\int <!--\; \int \; -->\frac{1}{2\sqrt{u}}du$
Factor out constants:
$\frac{1}{2}\int <!--\; \int \; -->\frac{1}{\sqrt{u}}du$
The integral of $\frac{1}{\sqrt{u}}$ is $2\sqrt{u}$:
$\sqrt{u}+\text{constant}$
Is this so? How can the antiderivative of $\frac{1}{\sqrt{u}}$ be $2\sqrt{u}$?

$$\begin{gathered} \frac{dy}{dx}=y^2+3x^2-<!--\; -\; -->1, \\ y(1)=1 \end{gathered}$$
on $D=|x-<!--\; -\; -->1|<=1,|y-<!--\; -\; -->1|<=1$
Find the second approximation to the solution and estimate the error term

What is the limit of the following function, as x approaches 1?
$\underset{x\to <!--\; \backslash rightarrow\; -->1}{lim}\frac{\ln <!--\; \; -->(x)}{\arctan <!--\; \; -->(1-<!--\; -\; -->x)}.$

Prove or disprove: The product of two functions has antiderivatives if one of the functions is continuous and the other has antiderivatives
$I\subseteq <!--\; \subseteq \; -->\mathbb{R}$ is an interval and $f,g:I\to <!--\; \to \; -->\mathbb{R}$.
f has antiderivatives (there exists $F:I\to <!--\; \to \; -->\mathbb{R}$ with $F^{\prime }=f$)
g is continuous on I
Prove that the product $f\cdot <!--\; \cdot \; -->g$ has antiderivatives
This is true when f is continuous or bounded, but I could neither prove it in the general case, nor find a counterexample

Find the complete solution of the given differential equation
$\frac{dy}{dx}=\frac{3x\sqrt{1+y^2}}{y}$
I know how to solve it if the right side didn't contain either x or y, but I can't think of a way to arrange them. I think I should solve for y, but I'm not sure. Could anyone show me the solution?

If $f\in <!--\; \in \; -->L_2(a,b)$, then ${\int <!--\; \int \; -->}_a^{x}f(y)dy\in <!--\; \in \; -->L_2(a,b)$?
If $f\in <!--\; \in \; -->L_2(a,b)$, then I want to show that the antiderivative $F(x):={\int <!--\; \int \; -->}_a^{x}f(y)dy$.
is in $L_2$ (I guess this is true). If $L_2(a,b)$ would be closed under pointwise product, i.e. if $f,g\in <!--\; \in \; -->L_2$, then $f\cdot <!--\; \cdot \; -->g\in <!--\; \in \; -->L_2$, then this would follow easily, but I guess this is not true. So any hints how to show the closure under taking antiderivatives?

Does the limit
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}\ln <!--\; \; -->\left(\frac{k(1+\mathrm{\alpha <!--\; \alpha \; -->})}{n-<!--\; -\; -->k+1+(n-<!--\; -\; -->k)\mathrm{\alpha <!--\; \alpha \; -->}}\right)$
exist?

Antiderivative Help
I am trying to find f(x) if I know that
$f^{″}(x)=-<!--\; -\; -->2+12x-<!--\; -\; -->12x^2,f(0)=4,f^{\prime }(0)=12$
First I found the first derivative $f^{\prime }(x)=-<!--\; -\; -->2x+6x^2-<!--\; -\; -->4x^3+C$.
and then I found the function, which is:
$f(x)=-<!--\; -\; -->x^2+2x^3-<!--\; -\; -->x^4+Cx+D$
Now I am lost as to what to do with those values they gave me $f(0)=4,f^{\prime }(0)=12$.

I wanted to prove $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->x}{x}=0$ by the squeeze theorem.

I want to find the limit of the following sequence
$x_n={\left(\frac{\log <!--\; \; -->(n+2)}{\log <!--\; \; -->(n+1)}\right)}^{n\log <!--\; \; -->n}$

Calculate the antiderivative of a given function
Consider the function $f:[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{4})$, $f(x)=\frac{(\cos <!--\; \; -->x+\sin <!--\; \; -->x{)}^n}{(\cos <!--\; \; -->x-<!--\; -\; -->\sin <!--\; \; -->x{)}^{n+2}}$, where $n\in <!--\; \in \; -->{\mathbb{N}}^{\ast <!--\; \ast \; -->}$. Find the antiderivative F of f such that $F(0)=\frac{1}{2(n+1)}$.
I've noticed that $(\cos <!--\; \; -->x-<!--\; -\; -->\sin <!--\; \; -->x{)}^{\prime }=-<!--\; -\; -->(\cos <!--\; \; -->x+\sin <!--\; \; -->x)$, but I can't figure out how to use this in finding F.

I am trying to solve an exercise to show that this function
$f(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-<!--\; -\; -->1}$

How to evaluate $\underset{x\to <!--\; \to \; -->0}{lim}\frac{\sqrt[n]{1+x}-<!--\; -\; -->1}{x},n\in <!--\; \in \; -->\mathbb{Z}$

I have the following function:
$\frac{\sqrt[3]{x+3}(\sqrt{x^2+1}-<!--\; -\; -->\sqrt{x^2+3x+4})}{\sqrt[6]{x^2+5}}$

Antiderivative of a function of the antiderivative
I've got $\frac{\mathrm{\partial <!--\; \partial \; -->}y(x(t),t)}{\mathrm{\partial <!--\; \partial \; -->}t}=c\ast <!--\; \ast \; -->(x(t)-<!--\; -\; -->y(x,t))$
Let's consider the general case where $\frac{\mathrm{\partial <!--\; \partial \; -->}y(t)}{\mathrm{\partial <!--\; \partial \; -->}t}=f(y(t))$
Given that the derivative depends on the original, how do I get the antiderivative?

Proof: There is no function $f\in <!--\; \in \; -->C^2({\mathbb{R}}^{\mathbb{3}})$ with gradient $\mathrm{\nabla <!--\; \nabla \; -->}f(x,y,z)=(yz,xz,x{y}^2)$
How can one show that there is no function, which is a continuously partially derivable function $f\in <!--\; \in \; -->C^2({\mathbb{R}}^{\mathbb{3}})$ with this gradient $\mathrm{\nabla <!--\; \nabla \; -->}f(x,y,z)=(yz,xz,x{y}^2)$.
I thought about using the Hessian matrix since one has to calculate all second partial derivatives of f there.
Since only the gradient is given, can I calculate the antiderivatives first:
$yz=xyz$
$xz=xyz$
$x{y}^2=x{y}^{2}z$
Now I want to calculate the antiderivatives of the antiderivatives:
$xyz=\frac{yz{x}^2}{2}$
$xyz=\frac{yz{x}^2}{2}$
$x{y}^{2}z={\displaystyle \frac{y^{2}z{x}^2}{2}}$
I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...