Primitive of Inverse Hyperbolic Cotangent Function/Proof 1

Theorem

$\ds \int \arcoth x \rd x = x \arcoth x + \frac {\map \ln {x^2 - 1} } 2 + C$

for $x^2 > 1$.

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:

$\ds u$

$=$

$\ds \arcoth x$

$\ds \leadsto \ \ $

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

$=$

$\ds \frac {-1} {x^2 - 1}$

and let:

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

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds x$

Then:

$\ds \int \arcoth x \rd x$

$=$

$\ds x \arcoth x - \int x \paren {\frac {-1} {x^2 - 1} } \rd x + C$

$\ds $

$=$

$\ds x \arcoth x + \int \frac {x \rd x} {x^2 - 1} + C$

$\ds $

$=$

$\ds x \arcoth x + \paren {\frac 1 2 \map \ln {x^2 - 1} } + C$

$\ds $

$=$

$\ds x \arcoth x + \frac {\map \ln {x^2 - 1} } 2 + C$

simplifying

$\blacksquare$