Master Calculus 1 Practice Problems with Expert Help

What is $\int <!--\; \int \; -->\frac{3}{4}t\cos <!--\; \; -->(5t)dt$?

What is $\int <!--\; \int \; -->\frac{2}{3}e^{-<!--\; -\; -->x}{x}^{2}dx$?

Find the integral of $\frac{3}{2}t\sin <!--\; \; -->(5t)$ with respect to t.

Find the integral of $t^7-<!--\; -\; -->4$ with respect to t.

Find the integral of $-<!--\; -\; -->x^2+3x-<!--\; -\; -->2$ with respect to x.

Find the integral of $\frac{7}{8}e^{2t}{t}^2$ with respect to t.

Find the integral of $\frac{2e^x{x}^2}{7}$ with respect to x.

What is $\int <!--\; \int \; -->(-<!--\; -\; -->x^2-<!--\; -\; -->2)dx$?

Antiderivative of periodic function
I practise before the real analysis exam and I got stuck on few excercises.
"if F is an antiderivative of a continuous periodic function and F is bounded, then F is periodic"
$\to <!--\; \to \; -->$ one should give a proof or counterexample.
(I'm convinced it's true, but how to prove it?)

Find the antiderivative of f
Let $S=\{0\}\cup <!--\; \cup \; -->\{\frac{1}{n}:n\in <!--\; \in \; -->\mathbb{N}\}$ and define $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$
a) Prove that $|f(x)-<!--\; -\; -->f(y)|\le <!--\; \le \; -->|x-<!--\; -\; -->y|$ for any $x,y\in <!--\; \in \; -->\mathbb{R}$.
b) Find the antiderivatives of f
I obtained the following form for the function:
$f(x)=\{\begin{array}{ll}-<!--\; -\; -->x,& x\le <!--\; \le \; -->0\\ x-<!--\; -\; -->\frac{1}{n+1},& \frac{1}{n+1}\le <!--\; \le \; -->x\le <!--\; \le \; -->\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n})\\ \frac{1}{n}-<!--\; -\; -->x,& \frac{1}{2}(\frac{1}{n+1}+\frac{1}{n})<x<\frac{1}{n}\\ x-<!--\; -\; -->1,& x\ge <!--\; \ge \; -->1\end{array}$
I then tried to prove the inequality from a) considering many cases for x and y, but there are many possibilities. For the antiderivatives, I tried to combine the antiderivatives for each of the 4 branches of the function, but again, the calculations are messy.

Antiderivatives in complex analysis
Let U be an open set in $\mathbb{C}$ and $f:U\to <!--\; \to \; -->E$ be holomorphic on U. Let now $\mathrm{\gamma <!--\; \gamma \; -->}$ be a path in U that begins with $z_1\in <!--\; \in \; -->U$ and ends with $z_2\in <!--\; \in \; -->U.$. I wonder for which assumptions the following formula ${\int <!--\; \int \; -->}_{\mathrm{\gamma <!--\; \gamma \; -->}}\frac{df}{dz}(z)=f(z_2)-<!--\; -\; -->f(z_1)$ is true?
I have found many counterexamples (such as $f^{\prime }(z)=\frac{1}{z}$) where the above formula fails if the set U is not simply connected. Here is a related result.
Another question, if f is well defined on $\stackrel{¯<!--\; ¯\; -->}{U}$ and $z_1\in <!--\; \in \; -->\mathrm{\partial <!--\; \partial \; -->}U$ is the above formula also true?

A function of $2$ variable is given by,
$f(x,y)=e^{2x-<!--\; -\; -->3y}$
How to find tangent approximation to $f(0.244,1.273)$ near $(0,0)?$?

Evaluating $\int <!--\; \int \; -->(\frac{3}{5}-<!--\; -\; -->\frac{8}{x})dx$
I am unsure how to get the antiderivative of $f(x)=\frac{3}{5}-\frac{8}{x}$
I know the answer is $-<!--\; -\; -->8\ln <!--\; \; -->(|x|)+C+\frac{3x}{5}$ as it says in my textbook.
But I am unsure how I get something like ln(x) in the answer? Could someone guide me through this? I am totally lost as for other antiderivatives I have no problem but this stuns me.

Solve over the reals for x: $$64=4^{7x+5}$$.

Solve over the reals for x: $$2^{4x+3}=32$$.

Solve over the reals for s in the equation $$32=2^{1-<!--\; -\; -->s}$$.

Find the integral of $$\frac{1}{3(6x-<!--\; -\; -->4{)}^2}$$ with respect to x.

What is $$\int <!--\; \int \; -->\frac{4}{5(4t+4)}dt$$?

Find the integral of $$\frac{1}{3}\cos <!--\; \; -->(4t+1)$$ with respect to t.

Antiderivative of a branch function
Let $f:\mathbb{R}\mapsto <!--\; \mapsto \; -->\mathbb{R}$, $f(x)=e^x$ for $x\le <!--\; \le \; -->0$ and $ax+b$ for $x>0$. I want to determine the real numbers a and b for which f admits antiderivatives.
I took a plausible antiderivative $F(x)=e^x+c$ for $x<0$, $F(0)=\mathrm{\alpha <!--\; \alpha \; -->}$ and $a\frac{x^2}{2}+bx+d$ for $x>0$ and using the fact that F must be first continuous and then differentiable, I showed that $b=1$.
This is easy to see as from continuity we must have that $d=\mathrm{\alpha <!--\; \alpha \; -->}=c+1$.
Then, since F is clearly differentiable on $\mathbb{R}\setminus <!--\; \setminus \; -->\{0\}$ if we impose the condition that the left and right derivatives at 0 of F to be equal so we must have:
$\underset{x\to <!--\; \to \; -->0,x<0}{lim}\frac{F(x)-<!--\; -\; -->F(0)}{x-<!--\; -\; -->0}=\underset{x\to <!--\; \to \; -->0,x>0}{lim}\frac{F(x)-<!--\; -\; -->F(0)}{x-<!--\; -\; -->0}$
Thus: $\underset{x\to <!--\; \to \; -->0,x<0}{lim}\frac{e^x+c-<!--\; -\; -->c-<!--\; -\; -->1}{x}=\underset{x\to <!--\; \to \; -->0,x>0}{lim}\frac{a\frac{x^2}{2}+bx+c+1-<!--\; -\; -->c-<!--\; -\; -->1}{x}$
Which gives us $b=1$.
My question is then, does this suffice? Can a be chosen arbitrarily such that f has antiderivatives as long as $b=1$. It would seem that the function $F(X)=e^x+c$ for $x\le <!--\; \le \; -->0$ and $F(x)=a\frac{x^2}{2}+x+c+1$ for $x>0$ for $c\in <!--\; \in \; -->\mathbb{R}$ does indeed satisfy the necessary conditions.