Primitive of Reciprocal of x squared minus a squared/Logarithm Form/Lemma

Lemma
Let $a \in \R_{>0}$ be a strictly positive real constant.

Then:
 * $\map \ln {\dfrac {x - a} {x + a} }$ is defined $\size x > a$


 * $\map \ln {\dfrac {a - x} {a + x} }$ is defined $\size x < a$

Proof
We have that the real natural logarithm is defined only on the strictly positive real numbers.

Hence:
 * $\map \ln {\dfrac {x - a} {x + a} }$ is defined $\dfrac {x - a} {x + a} > 0$
 * $\map \ln {\dfrac {a - x} {a + x} }$ is defined $\dfrac {a - x} {a + x} > 0$

First we note that if $\size x = a$, then either the numerator or denominator of the arguments of the logarithm functions in question are either $0$ or undefined.

Hence the expressions have no meaning unless $\size x \ne a$.

The following table indicates whether each of $x + a$, $a - x$ and $x - a$ are positive $(+)$ or negative $(-)$ on the domains in question.

$\begin {array} {c|ccc|cc} & x + a & a - x & x - a & \dfrac {x - a} {x + a} & \dfrac {a - x} {a + x} \\ \hline a < x     & + & - & + & + & - \\ 0 < x < a & + & + & - & - & + \\ -a < x < 0 & + & + & - & - & + \\ x < -a    & - & + & - & + & - \\ \end {array}$

Hence:
 * $\map \ln {\dfrac {x - a} {x + a} }$ is defined $x > a$ or $x < -a$
 * $\map \ln {\dfrac {a - x} {a + x} }$ is defined $-a < x < a$

as we were required to show.