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

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Theorem

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

Let $\size x > a$.

Then:

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


Proof

Let $\size x > a$.

Then:

\(\ds \int \frac {\d x} {x^2 - a^2}\) \(=\) \(\ds -\frac 1 a \arcoth \frac x a + C\) Primitive of $\dfrac 1 {x^2 - a^2}$: $\arcoth$ 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
\(\ds \) \(=\) \(\ds -\dfrac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {2 a} \map \ln {\frac {x - a} {x + a} } + C\) Reciprocal of Logarithm

$\blacksquare$


Sources