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I have puzzled over this for at least an hour, and have made little progress.

I tried letting $x^2 = \frac{1}{3}\tan\theta$, and got into a horrible muddle... Then I tried letting $u = x^2$, but still couldn't see any way to a solution. I am trying to calculate the length of the curve $y=x^3$ between $x=0$ and $x=1$ using

$L = \int_0^1 \sqrt{1+\left[\frac{dy}{dx}\right]^2} \, dx $

but it's not much good if I can't find $\int_0^1\sqrt{1+9x^4} \, dx$

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    parsing links from text is an arcane art; most early terminations can be fixed by adding url-encoding (in this case, `^` to `%5E`, `{` to `%7B`, `}` to `%7D`): http://wolframalpha.com/input/?i=Integrate[Sqrt[1%2B9x%5E4]%2C%7Bx%2C0%2C1%7D]2012-11-19

2 Answers 2

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If you set $x=\sqrt{\frac{\tan\theta}{3}}$ you have: $ I = \frac{1}{2\sqrt{3}}\int_{0}^{\arctan 3}\sin^{-1/2}(\theta)\,\cos^{-5/2}(\theta)\,d\theta, $ so, if you set $\theta=\arcsin(u)$, $ I = \frac{1}{2\sqrt{3}}\int_{0}^{\frac{3}{\sqrt{10}}} u^{-1/2} (1-u^2)^{-7/2} du,$ now, if you set $u=\sqrt{y}$, you have: $ I = \frac{1}{4\sqrt{3}}\int_{0}^{\frac{9}{10}} y^{-3/4}(1-y)^{-7/2}\,dy $ and this can be evaluated in terms of the incomplete Beta function.

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    @Jack D'Aurizio Shouldn't it be $-7/4$? And in order to use Beta Function we need b > 0, although b - 1 = -\frac{7}{2} \implies b = -\frac{5}{2} < 02015-11-17
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try letting $3x^2=\tan(\theta)$,

or alternatively $3x^2= \sinh(\theta)$.

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    I think I let $x^2 = \frac{1}{3}\tan(\theta)$, which is the same thing. (I put the wrong thing in my initial post, but I'll edit it now.)2012-11-20