When trying to calculate arc length, what is the easiest way to approach the $(dy/dx)^2$ portion? When trying to calculate arc length, what is the easiest way to approach the $(dy/dx)^2$ portion?
If I have:
$$x = \frac{1}{3}\sqrt{y}(y-3),\qquad 1\leq y\leq 9;$$
I take the derivative of the function and get $\frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2}$.
Next I have to square the derivative and I got
$$\frac{1}{4}y^{1/4} + \frac{1}{2} + \frac{1}{4}y^{-1}$$
after adding the 1 from the formula (for arc length) to it. 
Now to condense everything into the formula up to that point I would have:
$$L = \int_1^9 \sqrt{ \frac{1}{4}y^{1/4} + \frac{1}{2} + \frac{1}{4}y^{-1}}$$
Now in order to get rid of that radical I would have to get some sort of perfect square but the trouble is sometimes it's difficult to see it right away, and I don't really see it in this one. Is there a better way to go about these problems other than just "looking" at it and trying to figure it out?a
 A: Updated. I assume there is a typo in the title and in the first sentence of the question, as I commented, and that you want to evaluate $\displaystyle\int_{1}^{9}\sqrt{1+\left( \dfrac{dx}{dy}\right) ^{2}}\mathrm{d}y$, where $x=\dfrac{1}{3}\sqrt{y}\left( y-3\right) $. Its derivative is
$$\dfrac{\mathrm{d}x}{\mathrm{d}y}=\dfrac{1}{2}y^{1/2}-\dfrac{1}{2}y^{-1/2}.$$
So
$$\left( \frac{1}{2}y^{1/2}-\frac{1}{2}y^{-1/2}\right) ^{2}=\frac{1}{4}y-%
\frac{1}{2}+\frac{1}{4}y^{-1},$$
and
$$1+\left( \frac{1}{2}y^{1/2}-\frac{1}{2}y^{-1/2}\right) ^{2}=\frac{1}{4}y+%
\frac{1}{2}+\frac{1}{4}y^{-1}.$$

Now in order to get rid of that radical I would have to get some sort of perfect square but the trouble is sometimes it's difficult to see it right away, and I don't really see it in this one. Is there a better way to go about these problems other than just "looking" at it and trying to figure it out?

Hint: 
$$\frac{1}{4}y+\frac{1}{2}+\frac{1}{4}y^{-1}=\frac{1}{4}\frac{\left( y+1\right)
^{2}}{y},$$
or use the completing the square technique.
Note: most of the times inside the radical you have a function $f(y)$ which is not a perfect square nor anything similar. What you get as integrand is a radical   $R(y)=\sqrt{f(y)}$. And you have to integrate it using the normal integration techniques: substitution or by parts. But it is not guaranteed that the integral has a closed form. However, in the present case you do obtain a closed form. 
