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