# 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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– Patrick Da Silva Nov 19 '12 at 9:42

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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`): wolframalpha.com/input/… – ysth Nov 19 '12 at 9:51

## 2 Answers

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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Unfortunately I don't know anything about the incomplete Beta function, but thanks anyway. – daviewales Nov 20 '12 at 10:31

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.) – daviewales Nov 20 '12 at 10:28
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