<s>[INST] What is the next step in the parametric arc length problem? Find the length of the curve $x = \frac{1}{3}(2t^2 +1)^\frac{3}{2}$, $y = \frac{1}{2}t^2$ from 0 to 4.
I believe the arc length formula for a parametric equation is $\int_a^b \sqrt{ (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 } dt $
So, I'll need to get the derivative of both x:
$$ \frac{dx}{dt} = \frac{3}{2} (\frac{1}{3} )(2t^2 +1)^\frac{1}{2} $$
Simplified:
$$ \frac{1}{2} (2t^2 + 1)^\frac{1}{2} * (4t) $$
Square it (did I square it correctly?):
$$ 4t^2 (2t^2 + 1) $$
Multiply through:
$$ 8t^4 + 4t^2 $$
Derivative of y:
$$ y = t $$
Square it:
$$ y = t^2 $$
Plug it in:
$$ \int_0^4 \sqrt{ 8t^4 + 4t^2 + t^2 } $$
Add like terms
$$ \int_0^4 \sqrt{8t^4 + 5t^2} $$
Pull out $t^2$
$$ t \int_0^4 \sqrt{8t^2 + 5} $$
If everything is correct so far, what do I do next? 
 [/INST] Using the method of $u$-substitution let $u=8t^2+5$, then ${1\over 16}du=tdt$. When $t=0$, $u=5$ and when $t=4$, $u=133$. Thus $$\int_0^4 t\sqrt{8t^2+5}dt={1\over 16}\int_5^{133}\sqrt{u}du.$$ Integrating the right-hand side we obtain
$$\left.\left({1\over 24}u^{3\over 2}\right)\right|_5^{133}.$$ Which becomes $${1\over {24}}(133\sqrt{133}-5\sqrt5)\approx63.44.$$
</s>