Find the length of the curve y=ln[(e^x-1)/(e^x+1)] from x=1 to x=2?

1 Answer
Apr 14, 2018

# approx 1.127#

Explanation:

#y=ln[(e^x-1)/(e^x+1)] =ln[tanh (x/2)] #

#y' = 1/(tanh (x/2) ) sech^2 (x/2) * 1/2#

#= 1/(2 cosh (x/2) sinh (x/2) )= csch (x) #

#s = int_a^b sqrt(1 + (y'(x))^2) \ dx#

#= int_1^2 sqrt(1 + csch^2 x) \ dx#

# = int_1^2 coth x \ dx#

# = [ ln (sinh x) ]_1^2#

# = ln( (e^2 - 1/e^2)/(e - 1/e) ) approx 1.127#