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

Theorem

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

Then:

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

Corollary

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

Proof

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

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:

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

\(=\)

\(\ds \map \ln {\dfrac 1 {x + \sqrt {x^2 + a^2} } }\)

Logarithm of Reciprocal

\(\ds \)

\(=\)

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

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

\(\ds \)

\(=\)

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

Difference of Two Squares

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(=\)

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

Difference of Logarithms

$\blacksquare$

Also see

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