Primitive of Reciprocal of Root of x squared minus a squared/Inverse Hyperbolic Cosine Form

Theorem

 * $\ds \int \frac {\d x} {\sqrt {x^2 - a^2} } = \dfrac {\size x} x \arcosh {\size {\frac x a} } + C$

for $x^2 > a^2$.

Proof
When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.

When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined.

Hence the domain needs to be restricted to $\size x > a$, or that is: $\size {\dfrac x a} > 1$.

Let $x > a$.

Let:

Let $x < -a$.

Let $z = -x$.

Hence:
 * $\dfrac {\d z} {\d x} = -1$

Then:

Also see

 * Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$: Inverse Hyperbolic Sine Form
 * Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$