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I basically get it, down to tanh(x)=(e^(2x)-1)/(e^(2x)+1)

It is right isnt it?

Indeed, but how do you go from here to the inverse of tanh(x)?

do u know componendo - dividendo ?

Take tanh x=y. Then apply log on both sides.

Solve for x.

if a/b = c/d
then
(a+b) /(a-b) = (c+d)/ (c-d)
did u know this ?

That's what I did, I didn't get anywhere on the log expression though. Hmm, not sure hartnn.

that rule is called componendo-dividendo
apply that for
y/1= (e^(2x)-1)/(e^(2x)+1)

I'd have to give a proof in order to utilize it though. D:
It's supposed to be solved without it.

that will isolate e^2x
and hence you can apply log

solved without what ?

No shortcuts. :D

Algebraically

Great! Good work!

hint for next step
factor out e^2x from on left side

but did u get what i did ?

\(\huge \dfrac{1}{2}\ln (\dfrac{1+x}{1-x})\)
you should get this

-1-y / y-1 = (1+y)/ (1-y)
the 1/2 will come outside ln right ?
because
2x= ln (...)

What happens after you apply ln?

e^2x =[(1+y)/(1-y) ]
clear till here ??

Man I feel stupid. D:

Absolutely.

any more doubts?

Nope, thanks a lot! :)

welcome ^_^