# Primitive of Power of x by Inverse Hyperbolic Cosecant of x over a

## Theorem

$\ds \int x^m \arcsch \frac x a \rd x = \begin{cases} \ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a + \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x > 0 \\ \ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a - \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x < 0 \\ \end{cases}$

## Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$

let:

 $\ds u$ $=$ $\ds \arcsch \frac x a$ $\ds \leadsto \ \$ $\ds \frac {\rd u} {\rd x}$ $=$ $\ds \frac {-a} {\size x \sqrt{a^2 + x^2} }$ Derivative of $\arcsch \dfrac x a$

and let:

 $\ds \frac {\rd v} {\rd x}$ $=$ $\ds x^m$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds \frac {x^{m + 1} } {m + 1}$ Primitive of Power

Then:

 $\ds \int \frac {\arcsch \dfrac x a \rd x} {x^2}$ $=$ $\ds \paren {\arcsch \frac x a} \paren {\frac {x^{m + 1} } {m + 1} } - \int \paren {\frac {x^{m + 1} } {m + 1} } \paren {\frac {-a} {\size x \sqrt{a^2 + x^2} } } \rd x + C$ Integration by Parts $\ds$ $=$ $\ds \begin {cases} \ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a + \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x > 0 \\ \ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a - \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x < 0 \\ \end {cases}$ Definition of Absolute Value

$\blacksquare$