Calculus

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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 -

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 -

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 -

take 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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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?
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