How to get (Taylor) series formula for arcosh?

[swampwiz]

I'm looking at the series published @ Wikipedia: http://en.wikipedia.org/wiki/Inverse_hyperbolic_function

There is a series for arsinh, which I was able to derive with no problem - basically take the derivative of arsinh, which is a radical, then apply the general binomial expansion, which gives a series, from which successive derivatives can be determined, then do a Taylor series @ 0, which for every [ ( f⁽ⁿ⁾ / n! ) xⁿ ] term cancels out every term except the one that has the x power as 0, leaving only one term for each Taylor series term, resulting in the nice ordered series for arsinh. However, I can't seem to get something that appears to go in the direction of the series as presented @ Wikipedia.

It appears that somehow if ( 1 / x ) is used instead of ( x ), then there is some type of series that is similar to that of arsinh (i.e., in the way that the series for cos & sin are related) with a strange ln ( 2 x ) term as well. I figure that there must be some underlying identity between arsinh & arcosh, that somehow related arsinh( x ) with arcosh( 1 / x ) and ln ( 2 x ) but I can't find it anywhere.

 

[fresh_42]

You can use ${\displaystyle \operatorname {arsinh} (x)=\operatorname {sgn} (x)\cdot \operatorname {arcosh} \left({\sqrt{x^{2}+1}}\right)}$ and for ${\displaystyle x\geq 1}$ we have ${\displaystyle \operatorname {arcosh} (x)=\operatorname {arsinh} \left({\sqrt {x^{2}-1}}\right)}$.