How do you put a hyperbolic arctan into a TI 83 calculator?

Question

zepdrix · Answer

Umm so $\tanh x$ can be written with exponentials.
So I think $arctanh x$ can be written with logs.

I think this is what you get,$$\large arctanh x=\frac{1}{2}\frac{\ln(1+x)}{\ln(1-x)}$$

zepdrix · Answer

You might wanna google it just to make sure, I'm a little rusty on identities D:
Putting that into your calculator should get you what you need c:

anonymous · Answer

ahhhh i finally found it online thru the manuel... took awhile... but just in case you wondered, there is a tanh(-1), button under the catalog selection of the calculator... thanks for looking.

anonymous · Answer

i also found that formula you just mentioned, which worked too... but it was .5 ln (1+r/1-r)

zepdrix · Answer

TI's have the hyperbolic functions? nice!