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

Theorem

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

Let $x > 1$.

Then:

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

Proof

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

Hence for all $x$:

$-x + \sqrt {x^2 + a^2} > 0$

and so:

$x - \sqrt {x^2 + a^2} < 0$

Hence:

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

Then we have:

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

\(=\)

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

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

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \ln \size {x - \sqrt {x^2 + a^2} } - \map \ln {a^2}\)

$\blacksquare$

Also see

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