ln((x+1)/(x-1))

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Solution for ln((x+1)/(x-1)) equation: 


D( x )

(x+1)/(x-1) <= 0

x-1 = 0

(x+1)/(x-1) <= 0

(x+1)/(x-1) <= 0

( x+1 )

x+1 <= 0 // - 1

x <= -1

( x-1 )

x-1 <= 0 // + 1

x <= 1

(-oo:-1)(-1:1)(1:+oo)x+1-++x-1--+

x-1 = 0

x-1 = 0

x-1 = 0 // + 1

x = 1

x in (-oo:-1) U (1:+oo)

ln((x+1)/(x-1)) = 0

ln((x+1)/(x-1)) = ln(e^0)

(x+1)/(x-1) = e^0

(x+1)/(x-1)-e^0 = 0

(x+1)/(x-1)-1 = 0

(x+1)/(x-1)+(-1*(x-1))/(x-1) = 0

x-1*(x-1)+1 = 0

2 = 0

2/(x-1) = 0

2/(x-1) = 0 // * x-1

2 = 0

x belongs to the empty set

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