<s>[INST] find the length of the curve $y= \int_{-2}^x\sqrt{3t^4-1} \, dt$ I'm stuck for a while not sure how to continue, or if there's a mistake that I did that prevent me to continue.
$$y= \int_{-2}^x\sqrt{3t^4-1} \ dt \ , \ -2≤x≤-1$$  
$$ y = F(x) - F(-2) $$
$$ y' = f(x) - f(-2) $$
$$ y' = \sqrt{3x^4-1} - \sqrt{47} $$
$$ (y')^2 = 3x^4-1 -2 \sqrt{(3x^4-1)\ 47} + 47$$
$$ 1+(y')^2 = 3x^4 -2 \sqrt{(3x^4-1)\ 47} + 47$$
Now how can I integrate this:
$$ \int_{-2}^{-1} \sqrt{3x^4 -2 \sqrt{(3x^4-1)\ 47} + 47} $$
And I cant get it to $\int_{-2}^{-1} \sqrt{ [\sqrt{3x^4-1} \ - \sqrt{47}]^2}$
 [/INST] Notice by the FTC 
$$ y' = \sqrt{3x^4-1}$$
and Thus,
$$ (y')^2 +1 = 3x^4 $$
it follows that 
$$ \mathcal{L} = \int\limits_{-2}^{-1} \sqrt{3}x^2 dx = ... $$
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