Primitive of x by Inverse Hyperbolic Secant of x over a

Theorem

 * $\ds \int x \arsech \frac x a \rd x = \dfrac {x^2} 2 \arsech \dfrac x a - \dfrac {a \sqrt {a^2 - x^2} } 2 + C$

where $\arsech$ denotes the real area hyperbolic secant.

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 $x \arsinh \dfrac x a$


 * Primitive of $x \arcosh \dfrac x a$


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


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


 * Primitive of $x \arcsch \dfrac x a$