Primitive of Power of x by Inverse Hyperbolic Secant of x over a

Theorem

 * $\ds \int x^m \arsech \frac x a \rd x = \dfrac {x^{m + 1} } {m + 1} \arsech \dfrac x a + \dfrac 1 {m + 1} \int \dfrac {x^m} {\sqrt {a^2 - x^2} } \rd x + C$

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 $x^m \arsinh \dfrac x a$


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


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


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


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