Arcoth exercise to do. Beweise: arcoth (x) = 1/2 log (x+1/x-1) for (-1<x<1)

Question

arcoth (x) = 1/2 log (x+1/x-1) for (-1<x<1)

Hello I tried to prove this for all x but I failed... hopefully somebody could help me 

Thanks Sydney

Answer 1

y = COTH(x) = (e^{2·x} + 1)/(e^{2·x} - 1)

Subst. z = e^{2·x}

z = (y + 1)/(y - 1)

e^{2·x} = (y + 1)/(y - 1)

2·x = LN((y + 1)/(y - 1))

x = 1/2·LN((y + 1)/(y - 1))

Und kann es sein, dass du deine Angabe der Definitionsmenge eigentlich ausnehmen wolltest? Skizziere doch mal beide Graphen?

Answer 2

Hello

y=(e^x+e^{-x})/(e^x-e^{-x})

y*(e^x-e^{-x})=(e^x+e^{-x})

(y-1)*e^x=(y+1)*e^{-x}

 now use ln  and you are almost finished, but only for |x|>1

Gruß lul