Master Calculus 1 Practice Problems with Expert Help

Antiderivative of $\frac{dy}{dx}$
Find the antiderivative of $\frac{dy}{dx}$ = $e^{2x}-<!--\; -\; -->x$ and $y=5$ when $x=0$.
the antiderivative formula is $e^{kx}dx$ = $\frac{1}{k}{e}^{kx}+c$.
I did it, $\frac{dy}{dx}$ = $e^{2x}-<!--\; -\; -->x$ and $y=5$ when $x=0$.
$\frac{dy}{dx}=\frac{1}{2}{e}^{2x}+c$ but the right answer is $y=\frac{1}{2}(e^{2x}-<!--\; -\; -->x^2+9)$

Find $\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\pi <!--\; \pi \; -->}}{lim}\frac{\cos <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+x)}{x-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}$ without using L'Hôpital

Prove that $\underset{\mathrm{\alpha <!--\; \alpha \; -->}\to <!--\; \to \; -->0^{+}}{lim}\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->1}}{n^{\mathrm{\alpha <!--\; \alpha \; -->}}}=1/2$

Proving that $\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{x}^{1/n}=1$ at $x><!--\; >\; -->0$

How do you evaluate: ${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{\log <!--\; \; -->x}{e^x+e^{-<!--\; -\; -->x}+1}\mathrm{d}x$

All possible antiderivatives for Integration by Parts
Just had a quick inquiry with regards to the formula for Integration by Parts. If I'm not mistaken, the formula states that $\int <!--\; \int \; -->f^{\prime }(x)g(x)=f(x)g(x)-<!--\; -\; -->\int <!--\; \int \; -->f(x)g^{\prime }(x)$
However, in the case that I try to substitute an antiderivative with a valid constant, the formula does not appear to work. I attempted to use a simple product such as $7x(x^2)$, using 7x as f′(x) and $x^2$ as g(x) respectively. I found that using $\frac{7}{2}{x}^2+5$ for the antidervative for 7x does not work in the formula, while $\frac{7}{2}{x}^2$ without the constant does indeed work. I am sure that I am missing something, however, why do both of these solutions not work, even though both are valid antiderivatives?

I have the following limit.
$L=\underset{x\to <!--\; \backslash rightarrow\; -->0}{lim}(\frac{x+4}{x+1}-<!--\; -\; -->\frac{1}{\log <!--\; \; -->(x+1)})$

Most general antiderivative involving sec x
I'm stumped on how to get the most general antiderivative, F(x) , of $f(x)=e^x+3secx(tanx+secx)$.
First, I split the equation on addition, since $\int <!--\; \int \; -->[f(x)+g(x)]dx=\int <!--\; \int \; -->f(x)dx+\int <!--\; \int \; -->g(x)dx$
$F(x)=\int <!--\; \int \; -->e^{x}dx+3\int <!--\; \int \; -->secx(tanx+secx)dx$
Then I use the Substitution Rule. sec(x) seems like a good choice for u since it's the inner function of a composition and du will encompass the other x terms
$u=secx\Rightarrow <!--\; \Rightarrow \; -->du=(tanx+secx)dx$
Then I have $$\begin{gathered} F(x)=e^x+3\int <!--\; \int \; -->(u)du \\ F(x)=e^x+\frac{3u^2}{2}+C \\ F(x)=e^x+\frac{3se{c}^{2}x}{2}+C \end{gathered}$$
However, the answer key says the correct answer is $F(x)=e^x+3secx+3tanx+C$.
Can someone please point out where I've gone wrong here?

Integration by substitution, ${\int <!--\; \int \; -->}_1^{a}f(x^s)dx={\int <!--\; \int \; -->}_1^{a^s}f(x){\displaystyle \frac{1}{s{x}^{1-<!--\; -\; -->1/s}}}dx$

Evaluate ${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{\ln <!--\; \; -->x}{{(x^2+1)}^n}dx$

Find the quadratic approximation of:
$\frac{sec(t)}{\sqrt{1-<!--\; -\; -->t^2}}$
for $t\approx <!--\; \approx \; -->0$

Physics related antiderivatives problem
I was wondering if I could get a little help with a calculus related word problem:
A car braked with a constant deceleration of 5 meters per second squared for 60 meters before stopping. How fast was the car traveling when the brakes were applied? I can't use the definite integral, only the antiderivative.

How do you find the instantaneous rate of change of the function $f(x)=x^2+3x-<!--\; -\; -->11$ when x=2?

Simplifying Integral Expression w/ Antiderivative
I'm trying to simplify this formula here. For the most part, evaluating this expression is straightforward. However I am very confused about how to approach this last term in the expression. I want to simplify this part of the formula and break it down further if possible by coming up with an antiderivative formulation or something similar to make it more understandable, and to be able to evaluate it in terms of its unknowns. It's clearly related to the normal distribution but I'm unsure of how to simplify it in terms of the normal as well. How should I approach this? What would be the best way to evaluate this term?

How to calculate
$\underset{x\to <!--\; \to \; -->0}{lim}(\frac{1+\tan <!--\; \; -->x}{1+\sin <!--\; \; -->x}{)}^{\frac{1}{\sin <!--\; \; -->x}}$

Define a function f such that for $x\ne 0f(x)=cos(1/x)$ and $f(0)=1/2$. does it admits antiderivatives?

$\left\{x_n\right\}$ is a sequence and $x_{n+3}\le <!--\; \le \; -->\frac{1}{5}(x_{n+2}+3x_{n+1}+x_n)$ for all $n\ge <!--\; \ge \; -->1$. Prove that $\left\{x_n\right\}$ is convergent.

How do you find the instantaneous rate of change of the function $x^3+2x^2+x$ when x=1?

Determine: ${\int <!--\; \int \; -->}_{-<!--\; -\; -->1}^1\frac{1-<!--\; -\; -->x}{1+x}dx$

I need to prove the limit using the definition of limit
$\underset{x\to <!--\; \to \; -->c}{lim}f(x)=L\leftrightarrow <!--\; \leftrightarrow \; -->\mathrm{\forall <!--\; \forall \; -->}\mathrm{ϵ<!--\; ϵ\; -->}>0\mathrm{\exists <!--\; \exists \; -->}\mathrm{\delta <!--\; \delta \; -->}>0:0<|x-<!--\; -\; -->c|<\mathrm{\delta <!--\; \delta \; -->}\to <!--\; \to \; -->|f(x)-<!--\; -\; -->L|<\mathrm{ϵ<!--\; ϵ\; -->}$