Primitive of Inverse Hyperbolic Cosine of x over a over x squared

Theorem

 * $\ds \int \frac 1 {x^2} \arcosh \dfrac x a \rd x = -\frac 1 x \arcosh \dfrac x a + \frac 1 a \arcsec \size {\frac x a} + C$

where $\arcosh$ denotes the real area hyperbolic cosine.

Proof
With a view to expressing the primitive in the form:
 * $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

and let:

Then:

Also see

 * Primitive of $\dfrac {\arsinh \frac x a} {x^2}$


 * Primitive of $\dfrac {\artanh \frac x a} {x^2}$


 * Primitive of $\dfrac {\arcoth \frac x a} {x^2}$