Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2

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

Let $x \in \R$ such that $\size x \ne a$.


 * $\displaystyle \int \frac {\d x} {x^2 - a^2} = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C$

Proof
From the $1$st logarithm form:


 * $\displaystyle \int \frac {\d x} {x^2 - a^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C & : \size x < a\\

& \\ \dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C & : \size x > a \\ & \\ \text {undefined} & : \size x = a \end {cases}$

From Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma:


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


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

Let $\size x > a$.

Then $\map \ln {\dfrac {x - a} {x + a} }$ is defined.

We have that:

So the result holds for $\size x > a$.

Let $\size x < a$.

Then $\map \ln {\dfrac {a - x} {a + x} }$ is defined.

We have:

We have that:

The result follows.