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I am confused as to what is occurring in this step in an arc length problem: enter image description here

Could anyone take a stab at trying to explain it to me? thanks

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The problem???? – Matt Jul 6 '11 at 20:06
The 3/2, 2/3 obviously cancel. The reason for doing multiplication by $1$ in this strange way is to make the later process of integrating by substitution more transparent. A bad idea pedagogically, since it is unmotivated, mathematics as magic tricks. And the student rightly asks: How would I think of that? – André Nicolas Jul 6 '11 at 21:05
up vote 7 down vote accepted

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: $$\begin{align*} \int_1^8\sqrt{\frac{x^{2/3}+1}{x^{2/3}}}\,dx &= \int_1^8\frac{\sqrt{x^{2/3}+1}}{\sqrt{x^{2/3}}}\,dx\\ &= \int_1^8\frac{\sqrt{x^{2/3}+1}}{x^{1/3}}\,dx\\ &= \int_1^8\left(\sqrt{x^{2/3}+1}\right)\times\frac{3}{2}\times\frac{2}{3}\times\frac{1}{x^{1/3}}\,dx\\ &= \frac{3}{2}\int_1^8\sqrt{x^{2/3}+1}\left(\frac{2}{3x^{1/3}}\right)\,dx \end{align*}$$

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Wow! how was I supposed to know that? haha – Matt Jul 6 '11 at 19:58
@Matt: In all seriousness: note that the $\frac{3}{2}$ "outside" and the $\frac{2}{3}$ "inside" cancel out, so they don't matter. Then just notice that the $x^{1/3}$ in the denominator is exactly the square root of the $x^{2/3}$ you had in the denomminator when you started. – Arturo Magidin Jul 7 '11 at 6:36

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 $(x)^{\frac{2}{3} \times \frac{1}{2}} = x^{1/3}$.

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