Negative of Logarithm of x plus Root x squared minus a squared

Theorem

Let $x \in \R: \size x > 1$.

Let $x > 1$.

Then:

$-\map \ln {x + \sqrt {x^2 - a^2} } = \map \ln {x - \sqrt {x^2 - a^2} } - \map \ln {a^2}$

First we note that if $x > 1$ then $x + \sqrt {x^2 - a^2} > 0$.

Hence $\map \ln {x + \sqrt {x^2 - a^2} }$ is defined.

Then we have:

$\ds -\map \ln {x + \sqrt {x^2 - a^2} }$

$=$

$\ds \map \ln {\dfrac 1 {x + \sqrt {x^2 - a^2} } }$

$\ds $

$=$

$\ds \map \ln {\dfrac {x - \sqrt {x^2 - a^2} } {\paren {x + \sqrt {x^2 - a^2} } \paren {x - \sqrt {x^2 - a^2} } } }$

multiplying top and bottom by $x - \sqrt {x^2 - a^2}$

$\ds $

$=$

$\ds \map \ln {\dfrac {x - \sqrt {x^2 - a^2} } {x^2 - \paren {x^2 - a^2} } }$

$\ds $

$=$

$\ds \map \ln {\dfrac {x - \sqrt {x^2 - a^2} } {a^2} }$

simplifying

$\ds $

$=$

$\ds \map \ln {x - \sqrt {x^2 - a^2} } - \map \ln {a^2}$

$\blacksquare$