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

Theorem
Let $x \in \R$ be a real number.

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

Proof
We have that $\sqrt {x^2 + a^2} > x$ for all $x$.

Thus:
 * $x + \sqrt {x^2 + a^2} > 0$

and so $\map \ln {x + \sqrt {x^2 + a^2} }$ is defined for all $x$.

Then we have:

Also see

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