<s>[INST] Arc length of $y^3 = x^2$ I am trying to find the arc length of $y^3 = x^2$ and I am suppose to use two formulas, one for in terms of x and one for in terms of y.
At first I need to find (0,0) (1,1) and I start with in terms of x
$$y\prime = \frac{2}{3}x^\frac{-1}{3}$$
$$\left(\frac{2}{3}x^\frac{-1}{3}\right)^2 = \frac{4}{9x^\frac{2}{3}}$$
$$\int_0^1 \sqrt{1 + \left( \frac{2}{3}x^\frac{-1}{3}\right)^2}\, dx$$
$$\int_0^1 \sqrt{1 + \frac{4}{9x^\frac{2}{3}}}\, dx$$
This is undefined at 0 so it is an improper function. I do not think I can continue.
Now in terms of y.
$$x = y^\frac{3}{2}$$
$$x \prime = \frac{3\sqrt{y}}{2}$$
$$ \left(\frac{3\sqrt{y}}{2}\right)^2 = \frac{9y}{4}$$
$$\int_0^1 \sqrt{1+ \frac{9y}{4}}\, dy$$
I have no idea how to factor that out, I have tried many ways but I can not get it.
 [/INST] For the last integral, let $u=1+\frac{9y}{4}$. Then $du=\frac{9}{4}\,dy$, so $dy=\frac{4}{9}\,du$. The indefinite integral is $\int \frac{9}{4}u^{1/2}\,du$, and it's almost over.
The first integral is a bit more scary. I would rewrite your $\sqrt{1 + \frac{4}{9x^{2/3}}}$ as  $\sqrt{\frac{9x^{2/3} + 4}{9x^{2/3}}}$ and then pull the denominator out to conclude that we are interested in the indefinite integral 
$$\int \frac{1}{3x^{1/3}}\sqrt{9x^{2/3}+4}\,\,dx$$
  Now let $u=9x^{2/3}+4$. Then $du=6x^{-1/3}\,dx=\frac{6}{x^{1/3}}\,dx$. Then the $\frac{1}{3x^{1/3}}\,dx$ term is just $\frac{1}{18}du$.   
You are right in viewing the above integral as an improper integral, since there is the appearance of trouble at $0$. In principle, we should integrate from $\epsilon$ to $1$, and find the limit as $\epsilon$ approaches $0$ from the right. In practice, the substitution $u=x^{2/3}$ makes the problem disappear.
Remark: There is a fair bit of difference in appearance of difficulty between the two approaches. But in fact they are just a substitution away from each other. However, the appropriate substitution may not be obvious.  
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