<s>[INST] Arc Length Problem: Assimilation of A Constant The problem I am working on is,

I understand the motivation of the step, but I am not quite sure how the author the solution manual absorbs the constant into the other term. What is the process?
EDIT:
Here is another one:

Here is my attempt to try and figure out what they are doing:
$1+[y']^2=[\frac{1}{2}(e^x-e^-x)]+1$
$1+[y']^2=[\frac{1}{2}(e^x-e^-x)+1]^2$
$1+[y']^2=[\frac{1}{2}(e^x-e^-x+2)]^2$ This is where I get stuck, because I can't truly see any sort of manipulation that would get rid of the 2 and put a plus sign between the exponential functions.
 [/INST] $\quad\quad\quad$First question: $$1 + \frac14\left(x^8 - 2 +\frac{1}{x^8}\right)\tag{1.1}$$
$$=\frac14\cdot 4 + \frac14\left(x^8 - 2 + +\frac{1}{x^8}\right)\quad=\quad\frac14\left(4 + x^8 -2 +\frac{1}{x^8}\right)\tag{1.2}$$
$$=\frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$$
$$=\frac14\left(x^4 + \frac{1}{x^4}\right)^2\tag{1.4}$$
$$(1.3) \to (1.4):\quad\quad\text{Note that}$$
$$\frac14 \left(x^4 + \frac{1}{x^4}\right)^2\tag{1.4}$$ $$= \frac14\left(x^{2\cdot 4} + 2\frac{x^4}{x^4} + \frac{1}{x^{2\cdot 4}}\right) = \frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$$


For your second question: in your work you have:
$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})]+1\tag{a}$$
$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})+1]^2\tag{b}$$
$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x}+2)]^2\tag{c}$$ 
How did you get from $(a)\to(b)$? Did you forget the exponent in $(a)$? Shouldn't $(a)$ be $$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})]^2+1\quad?$$And if so, you cannot bring the constant term $1$ into an expression that is exponentiated without first making appropriate computations on the exponentiation expression. Failing that, your move from $(b)\to (c)$ is affected.
In this case, first expand $\left[\dfrac{1}{2}(e^x-e^{-x})\right]^2$, then worry about adding the $1$ term:
$$1+[y']^2=\left[\frac{1}{2}(e^x-e^{-x})\right]^2+1\tag{2.1}$$
$$=\frac14\left(e^{2x} -2\frac{e^{2x}}{e^{2x}} + \frac{1}{e^{2x}}\right) + 1\tag{2.2}$$
$$=\frac14\left(e^{2x} -2 + \frac{1}{e^{2x}}\right) + \frac14\cdot4
\quad=\quad\frac14\left(e^{2x} -2 +\frac{1}{e^{2x}}+4\right)\tag{2.3}$$ $$= \left(\frac12\right)^2\left(e^{2x} +2 +\frac{1}{e^{2x}}\right)\quad=\quad\left[\frac12(e^x + e^{-x})\right]^2\tag{2.4}$$
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