# Integrate $\sqrt{1+9x^4} \, dx$

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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Doesn't look easy, here's what WolframAlpha pops out : wolframalpha.com/input/?i=Integrate[Sqrt[1%2B9x^4]%2C{x%2C0%2C1}] (you should copy the whole link, not just click on it.. for some reason it won't get all linked) but I guess maybe there's another way around, I really, really didn't think about it much. – Patrick Da Silva Nov 19 '12 at 9:42
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): wolframalpha.com/input/… – ysth Nov 19 '12 at 9:51

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.
@Jack D'Aurizio Shouldn't it be $-7/4$? And in order to use Beta Function we need $b > 0$, although $$b - 1 = -\frac{7}{2} \implies b = -\frac{5}{2} < 0$$ – Aaron Maroja Nov 17 '15 at 23:58
try letting $3x^2=\tan(\theta)$,
or alternatively $3x^2= \sinh(\theta)$.
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.) – daviewales Nov 20 '12 at 10:28