Arc length problem with a thorny integration

aucuneidee
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Homework Statement



So, the question gives a particle traveling over a path \gamma, and I need the arc length.

Homework Equations



The path is \gamma(t) : [1,4] \to ℝ^3, t \mapsto (t^2/2, t, ln(2t)).

We want the arc length over 1 \le t \le 4.

The Attempt at a Solution



First, the speed differential: ds = \left\| \gamma'(t) \right\| dt = \sqrt{t^2 + 1 + 1 /t^2} dt

Now, the arc length. \ell = \int_\gamma ds = \int_1^4 \sqrt{t^2 + 1 + 1 /t^2}dt.

But that's where the fun ends. I've tried a bunch of trig substitutions (e.g. t=\tan u), to no avail.

I also tried Wolfram online integrator, which returned a mess of symbols -- this problem should have a (reasonably) simple analytic solution.

Any ideas, anyone? I'd really appreciate any help!
 
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Didn't you notice that t^2+ 1+ 1/t^2= t^2+ 2+ 1/t^2- 1= (t+ 1/t)^2- 1?
 
HallsofIvy said:
Didn't you notice that t^2+ 1+ 1/t^2= t^2+ 2+ 1/t^2- 1= (t+ 1/t)^2- 1?

I had, but didn't realize it would help. I'll play around and see what I can come up with. :)

Thankyou for the quick response!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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