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 $$\int_0^1\sqrt{1+9x^4} \, dx$$

try letting $3x^2=\tan(\theta)$,
or alternatively $3x^2= \sinh(\theta)$.
