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December 18, 2014

December 18, 2014

Posted by **MS** on Thursday, September 12, 2013 at 4:20am.

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 4:37amsubstitute

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 5:44amI 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 2:54pmtake a trip on over to wolframalpha and enter

integral sqrt(x+1)/x dx

and then click on the "Show Step-by-Step 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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