Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2/Proof 3

Theorem

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

Proof

From the $1$st logarithm form:

$\ds \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 if and only if $\size x > a$

$\map \ln {\dfrac {a - x} {a + x} }$ is defined if and only if $\size x < a$

Let $\size x > a$.

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

We have that:

\(\ds \dfrac {x - a} {x + a}\)

\(>\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {x - a} {x + a}\)

\(=\)

\(\ds \size {\dfrac {x - a} {x + a} }\)

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:

\(\ds \dfrac {a - x} {a + x}\)

\(=\)

\(\ds -\dfrac {x - a} {x + a}\)

\(\ds \)

\(<\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {a - x} {a + x}\)

\(=\)

\(\ds \size {\dfrac {x - a} {x + a} }\)

The result follows.

$\blacksquare$