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

Question

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

Answer 1

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.

Answer 2

try letting $3x^2=\tan(\theta)$,

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