Integration of $\int\sqrt{\frac{a+x}{a-x}} dx$, exercise $45$ in section $6.22$ of Tom Apostol's Calculus Vol $1$ While looking at the solution here:
https://www.stumblingrobot.com/2015/12/01/evaluate-the-integral-of-axa-x12/

Evaluate the following integral for $a>0$.
$$\newcommand{\dd}{\; \mathrm{d}} \int\sqrt{\frac{a+x}{a-x}} \dd x.$$
First, we multiply the numerator and denominator in the fraction inside the square root by $a+x$ and do some simplification,
\begin{align*}   \int \sqrt{\frac{a+x}{a-x}} \, dx &= \int \sqrt{\frac{(a+x)^2}{(a-x)(a+x)}} \, dx \\[9pt]  &= \int \frac{a+x}{\sqrt{a^2-x^2}} \, dx \\[9pt]  &= \int \frac{a}{\sqrt{a^2-x^2}} \,dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx \\[9pt]  &= \int \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \, dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx. \end{align*}
Now, we evaluate these two integrals separately. ...

I wondered why it is fine to multiply the numerator and denominator by $(a+x)$?
In the original integrand, the square root has to be non-negative and the denominator has to be non-0, so $x \in [-a, a)$. Therefore, the end point $-a$ is still in the domain of the integrand, but it is not in the domain of the new integrand when we extend it with $(a+x)/(a+x)$, so I'm not sure why exactly that is justified.
If that's justified, please explain why. If that's not justified, please provide a full correct solution.
 A: $\sqrt{\frac{a+x}{a-x}}$, $a>0$, implies $\frac{a+x}{a-x} \geq 0$
So we have, $\frac{a+x}{a-x}=0$, implying $a+x=0$, which is trivial and uninteresting since $\int 0 dx=C$. Meaning, this is a special case for some particular value of $x$.
Or, we have $b>\frac{a+x}{a-x}>0$ for some $b\in \mathbb R^{+}$ which implies $b(a-x)>(a+x)>0$ which implies $a>-x$
This is not limited to end-points, we can generalize it to an arbitraty discontinuity that is isolated, given one trivial assumption that $F(x)$ is defined for $x=-a$, consider the second FTC:
Given a continous, integrable, function $f(x)$ on some interval $[a,b]$,  $\int_a^bf(x)=F(b)-F(a)$, given that $F'(x)=f(x)$.
In this case, suppose our function g(x) is continous, and integrable, on $[a,b]\setminus \{c\}$ which means $g(x)$ is integrable on $[a, c- \epsilon ] \cup [c + \epsilon ,b]$.
We get, $$\lim \limits_{d^{-} \to c}\int_a^dg(x)dx + \lim \limits_{e^{+} \to c}\int_e^b g(x)dx = \lim \limits_{d^{-}\to c}G(d)-G(a)+G(b)-\lim \limits_{e^{+}\to c}G(e)=G(c)-G(a)+G(b)-G(c)=G(b)-G(a)$$
The assumption here is that the two limits exist and are equal, which is guaranteed by the fact that $a=-x$ is a special case for some particular value of $x$. If you integrate the the function, $\sqrt{\frac{a+x}{a-x}}$, under the assumption that $x\neq -a$ and check the Limit, $x\to -a$ you can see that it exists, and is one and the same from both directions, then you can discharge the assumption.
A: There's a theorem that goes:$\quad$If $A{\cap}B$ has measure zero,$\quad$$f:A{\cup}B\to\mathbb R,$$\quad$and $f$ is integrable over each of $A,B$ and $A{\cap}B,$$\quad$then $\int_{A{\cup}B}f=\int_Af+\int_Bf.$
To justify excluding $-a$ from the given solution, we can apply the above theorem, letting $A=\{-a\}$ and $B=(-a,p)$ for each $p$ smaller than $a.$
