Primitive of Reciprocal of x by Root of x squared plus a squared/Inverse Hyperbolic Cosecant Form

Theorem

$\ds \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \csch^{-1} {\frac x a} + C$

Let:

$\ds u$

$=$

$\ds \csch^{-1} {\frac x a}$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds a \csch u$

$\ds \leadsto \ \ $

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

$=$

$\ds -a \csch u \coth u$

$\ds \leadsto \ \ $

$\ds \int \frac {\d x} {x \sqrt {x^2 + a^2} }$

$=$

$\ds \int \frac {-a \csch u \coth u} {a \csch u \sqrt {a^2 \csch^2 u + a^2} } \rd u$

$\ds $

$=$

$\ds -\frac a {a^2} \int \frac {\csch u \coth u} {\csch u \sqrt {\csch^2 u - 1} } \rd u$

$\ds $

$=$

$\ds -\frac 1 a \int \frac {\csch u \coth u} {\csch u \coth u} \rd u$

$\ds $

$=$

$\ds -\frac 1 a \int 1 \rd u$

$\ds $

$=$

$\ds -\frac 1 a u + C$

$\ds $

$=$

$\ds \frac 1 a \csch^{-1} {\frac x a} + C$

Definition of $u$

$\blacksquare$