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?

To integrate the expression ∫[(sqrt(1+x^2))/x] dx, we can use a substitution. Let's substitute u = 1 + x^2.

Differentiating both sides with respect to x, we get du/dx = 2x, or dx = du/(2x).

Now we need to substitute u and dx into the integral:

∫[(sqrt(1+x^2))/x] dx = ∫[(sqrt(u))/(x*2x)] du

Simplifying, we have:

∫[(sqrt(u))/(2x^2)] du

Next, we can simplify x^2 in terms of u by noticing that x^2 = u - 1:

∫[(sqrt(u))/(2(u-1))] du

Now we can separate this integrand into two parts:

∫[(sqrt(u))/2(u-1)] du = (∫[(sqrt(u))/(u-1)] du)/2

To integrate the remaining expression, we can use a trigonometric substitution. Let's substitute u - 1 = t^2, which implies du = 2t dt:

∫[(sqrt(u))/(u-1)] du = ∫[(sqrt(t^2+1))/t] 2t dt

Simplifying, we obtain:

2 ∫[(sqrt(t^2+1))] dt

Now, the integral becomes a standard integral of the form ∫(sqrt(a^2 + x^2)) dx, which evaluates to (1/2)(x*sqrt(a^2 + x^2) + a^2*sinh^(-1)(x/a)).

Applying this formula to our integral, where a = 1, we have:

2 ∫[(sqrt(t^2+1))] dt = t*sqrt(t^2+1) + sinh^(-1)(t)

Substituting t = sqrt(u-1), we get:

= sqrt(u-1)*sqrt(u) + sinh^(-1)(sqrt(u-1))

Finally, substituting u = 1 + x^2, we have:

∫[(sqrt(u))/(u-1)] du = sqrt(x^2)*sqrt(1+x^2) + sinh^(-1)(sqrt(x^2)) - sinh^(-1)(sqrt(1))

Evaluating this from x = 1 to x = 2, we have:

Arc length = ∫[(sqrt(1+x^2))/x] dx from x = 1 to x = 2
= [sqrt(x^2)*sqrt(1+x^2) + sinh^(-1)(x)] from x = 1 to x = 2
= [sqrt(1+x^2) + sinh^(-1)(x)] from x = 1 to x = 2