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

Theorem

 * $\ds \int x^2 \arcosh \frac x a \rd x = \dfrac {x^3} 3 \arcosh \dfrac x a - \dfrac {\paren {x^2 + 2 a^2} \sqrt {x^2 - a^2} } 9 + C$

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 $x^2 \arsinh \dfrac x a$


 * Primitive of $x^2 \artanh \dfrac x a$


 * Primitive of $x^2 \arcoth \dfrac x a$