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