What is going on in this step? (from arc length problem) I am confused as to what is occurring in this step in an arc length problem:

Could anyone take a stab at trying to explain it to me? thanks
 A: First, the square root of the quotient is the quotient of the square roots:
$$\sqrt{\frac{x^{2/3}+1}{x^{2/3}}} = \frac{\sqrt{x^{2/3}+1}}{\sqrt{x^{2/3}}}.$$
Next, the square root in the denominator simplifies with the exponent, since:
$$\sqrt{x^{2/3}} = \left(x^{2/3}\right)^{1/2} = x^{1/3}.$$
Next, introduce a factor of $1$, "disguised" as $\frac{3}{2}\times\frac{2}{3}$; finally, pull one of the two factors out of the integral, since it is constant. Thus:
$$\begin{align*}
\int_1^8\sqrt{\frac{x^{2/3}+1}{x^{2/3}}}\,dx &= \int_1^8\frac{\sqrt{x^{2/3}+1}}{\sqrt{x^{2/3}}}\,dx\\
&= \int_1^8\frac{\sqrt{x^{2/3}+1}}{x^{1/3}}\,dx\\
&= \int_1^8\left(\sqrt{x^{2/3}+1}\right)\times\frac{3}{2}\times\frac{2}{3}\times\frac{1}{x^{1/3}}\,dx\\
&= \frac{3}{2}\int_1^8\sqrt{x^{2/3}+1}\left(\frac{2}{3x^{1/3}}\right)\,dx
\end{align*}$$
A: In general if you have something like $\frac{a+b}{a}$ then you can write it as $$\frac{a+b}{a} = \frac{a}{a} + \frac{b}{a} = 1 + \frac{b}{a}$$
Next, since your $x^{2/3}$ has a square root, therefore, you have $(x)^{\frac{2}{3} \times \frac{1}{2}} = x^{1/3}$.
