Posted by MS on Thursday, September 12, 2013 at 4:20am.
Find arc length of y=logx from x=1 to x=2.
dy/dx)^2=1/x^2
arc length=Int of [sqrt(1+1/x^2)]dx
=Int of [sqrt(1+x^2)/x^2]
=Int of [sqrt(1+x^2)]/x from x=1 to x=2.
How to proceed further to integrate?

Calculus  Steve, Thursday, September 12, 2013 at 4:37am
substitute
x = tan u
dx = sec^2 u du
1+x^2 = sec^2 u
and you will wind up with some nice integrands involving tan u and sec u

Calculus  MS, Thursday, September 12, 2013 at 5:44am
I reached upto Int csc u sec^2 u du from u=arctan 1 to 2 but am not clear how to go further.

Calculus  Steve, Thursday, September 12, 2013 at 2:54pm
take a trip on over to wolframalpha and enter
integral sqrt(x+1)/x dx
and then click on the "Show StepbyStep Solution" button (you may have to register first)
and it will show all the intricacies of the substitution.
Or, recall that sec^2 = 1+tan^2.
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