# What is the arclength of #(t^2-lnt,lnt)# on #t in [1,2]#?

Arclength is given by:

Simplify:

Rearrange:

Factorize:

Integration is distributive:

Apply the double-angle Trigonometric identities:

Factorize the denominator:

Apply partial fraction decomposition:

Integrate directly:

Simplify:

Insert the limits of integration:

Simplify:

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To find the arc length of the curve ( (t^2 - \ln(t), \ln(t)) ) on the interval ([1, 2]), you use the formula for arc length: [ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} , dt ]

First, find the derivatives of (x) and (y) with respect to (t): [ \frac{dx}{dt} = 2t - \frac{1}{t} ] [ \frac{dy}{dt} = \frac{1}{t} ]

Now, compute the integrand: [ \sqrt{\left(2t - \frac{1}{t}\right)^2 + \left(\frac{1}{t}\right)^2} ]

Simplify the expression inside the square root: [ \sqrt{4t^2 - 4 + \frac{1}{t^2} + \frac{1}{t^2}} ] [ = \sqrt{4t^2 + \frac{2}{t^2} - 4} ]

Now integrate this expression from (t = 1) to (t = 2): [ L = \int_{1}^{2} \sqrt{4t^2 + \frac{2}{t^2} - 4} , dt ]

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