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

Theorem

Let $\size x > a$.

Then:

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

Proof

Let $\size x > a$.

Then:

\(\ds \int \frac {\d x} {a^2 - x^2}\)

\(=\)

\(\ds \frac 1 a \arcoth {\frac x a} + C\)

Primitive of $\dfrac 1 {a^2 - x^2}$: $\arcoth$ 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

$\blacksquare$