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

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Answer 1 · 2018-05-30T20:41:09

#= ln ((e^4 - 1)/(e^3 - e))#

# y=ln( (e^x-1)/(e^x+1))#

# =ln( tanh (x/2) )#

Arc length:

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

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

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

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

#= int_1^2 coth x dx#

#= ln ( sinh x ) |_1^2 #

#= ln ( sinh 2 ) - ln ( sinh 1 )#

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

#= ln ((e^4 - 1)/(e^3 - e))#