Re: [Axiom-developer] please check your Schaums

[Doug Stewart]

address@hidden wrote:
> In 14.661 Schaums claims:
>
> integral acoth(x/a) = x*acoth(x)+a/2*log(x^2-a^2)
>                         ^^^^^^^^
>
> Axiom claims
>
> integral acoth(x/a) = x*acoth(x/a)+a/2*log(x^2-a^2)
>                         ^^^^^^^^^^
>
> Is this a Schaums typo?
>
> Tim
>
>
>
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>   
My schaums is the same as your Schaums but it is old (not as old as 
yours) I have a New Schaums at work But I will not Be in to work today  :-)


My Maxima agrees with Axium.

integrate(acoth(x/a),x);
(%o8) (a*log(x^2/a^2-1))/2+x*acoth(x/a)