Primitive of Inverse Hyperbolic Cosecant of x over a

Theorem

$\ds \int \arcsch \frac x a \rd x = \begin {cases} x \arcsch \dfrac x a + a \arsinh \dfrac x a + C & : x > 0 \\ x \arcsch \dfrac x a - a \arsinh \dfrac x a + C & : x < 0 \end {cases}$

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:

$\ds u$

$=$

$\ds \arcsch \frac x a$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac {-a} {\size x \sqrt {a^2 + x^2} }$

Derivative of $\arcsch \dfrac x a$

and let:

$\ds \frac {\d v} {\d x}$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds x$

Primitive of Constant

Then:

$\ds \int \arcsch \frac x a \rd x$

$=$

$\ds x \arcsch \frac x a - \int x \paren {\frac {-a} {\size x \sqrt {a^2 + x^2} } } \rd x + C$

Integration by Parts

$\ds $

$=$

$\ds x \arcsch \frac x a + a \int \frac {x \rd x} {\size x \sqrt {a^2 + x^2} } + C$

Primitive of Constant Multiple of Function

$\ds $

$=$

$\ds x \arcsch \frac x a \begin {cases} \mathop + a \ds \int \dfrac {\d x} {\sqrt {a^2 + x^2} } + C & : x > 0 \\ \mathop - a \ds \int \dfrac {\d x} {\sqrt {a^2 + x^2} } + C & : x < 0 \end {cases}$

Definition of Absolute Value

$\ds $

$=$

$\ds \begin {cases} x \arcsch \dfrac x a + a \arsinh \dfrac x a + C & : x > 0 \\ x \arcsch \dfrac x a - a \arsinh \dfrac x a + C & : x < 0 \end {cases}$

Primitive of $\dfrac 1 {\sqrt {a^2 + x^2} }$

$\blacksquare$