Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x greater than a/Proof 2

Proof
Let $\size x > a$.

Then:

If $x > a$, then both $x + a > 0$ and $x - a > 0$.

So $\dfrac {x + a} {x - a} > 0$ and so:
 * $\ln \size {\dfrac {x + a} {x - a} } = \map \ln {\dfrac {x + a} {x - a} }$

If $x < -a$, then both $x + a < 0$ and $x - a < 0$.

So again $\dfrac {x + a} {x - a} > 0$ and so:
 * $\ln \size {\dfrac {x + a} {x - a} } = \map \ln {\dfrac {x + a} {x - a} }$

Hence the result.