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

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Theorem

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

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


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


Proof 1

Let $x > a$.

Then:

\(\ds \int \frac {\d x} {x^2 - a^2}\) \(=\) \(\ds -\frac 1 a \arcoth {\frac x a} + C\) Primitive of Reciprocal of $x^2 - a^2$ in $\arcoth$ form
\(\ds \) \(=\) \(\ds -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} } } + C\) $\arcoth {\dfrac x a}$ in Logarithm Form
\(\ds \) \(=\) \(\ds -\frac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \map \ln {\frac {x - a} {x + a} } + C\) Logarithm of Reciprocal
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C\) as $\dfrac {x - a} {x + a} > 0$ for $x > a$


Let $x < -a$.

Let $z = -x$.

Then:

$\d x = -\d z$

and we then have:

\(\ds \int \frac {\d x} {x^2 - a^2}\) \(=\) \(\ds \int \frac {-\d z} {\paren {-z}^2 - a^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\int \frac {\d z} {z^2 - a^2}\) simplifying
\(\ds \) \(=\) \(\ds -\frac 1 {2 a} \map \ln {\frac {z - a} {z + a} } + C\) from above
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \map \ln {\frac {z + a} {z - a} } + C\) Logarithm of Reciprocal
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \map \ln {\frac {-x + a} {-x - a} } + C\) substituting $-x$ back for $z$
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \map \ln {\frac {\size {x - a} } {\size {x + a} } } + C\) as $x < -a$
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C\)

The result follows.

$\blacksquare$


Proof 2

\(\ds \int \frac {\d x} {x^2 - a^2}\) \(=\) \(\ds \int \frac {\d x} {\paren {x - a} \paren {x + a} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \int \frac {\d x} {2 a \paren {x - a} } - \int \frac {\d x} {2 a \paren {x + a} }\) Partial Fraction Expansion
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\d x} {x - a} - \frac 1 {2 a} \int \frac {\d x} {x + a}\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \ln \size {x - a} - \frac 1 {2 a} \ln \size {x + a} + C\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \dfrac 1 {2 a} \ln \size {\dfrac {x - a} {x + a} } + C\) Difference of Logarithms

$\blacksquare$


Proof 3

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.

$\blacksquare$


Sources

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