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Long story short, I am trying to find the arc length of $y = 1/x + 1/(12x^3)$ on the interval $[1,2]$

The only part I am stuck on is how do I convert(my integrand): $(x^2/4)^2 + (1/x^2)^2 + 1/2$

to: $((x^2/4) + (1/x^2))^2$

This (my integrand) is under a square root, so I want to understand how I get this form.

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    By multiplying $(x^2/4)$ and $(1/x^2)$ we get a number, the variables reduces. This is good indication that $(a+b)^2=a^2+\color{red}{2ab}+b^2$ is useful. – 2017-02-23

2 Answers 2

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Complete the square: $$\begin{align}\left(\frac{x^2}4\right)^2+\left(\frac{1}{x^2}\right)^2 + \frac 12 &= \left(\frac{x^2}4\right)^2+\left(\frac{1}{x^2}\right)^2 + 2 \times \left(\frac{x^2}4\right) \left(\frac{1}{x^2}\right) \\ &= a^2 + b^2 + 2ab\end{align}$$ If you let $a = \frac{x^2}4$ and $b = \frac{1}{x^2}$. Recognise $a^2 + b^2 + 2ab = (a+b)^2$, then

$$\begin{align}\left(\frac{x^2}4\right)^2+\left(\frac{1}{x^2}\right)^2 + \frac 12 &= (a+b)^2 \\ &= \left(\frac{x^2}4+\frac{1}{x^2}\right)^2\end{align}$$

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It is easiest to factor out a $1/x^4$ from the expression to begin with. In particular, $$ (x^2/4)^2+(1/x^2)^2+1/2=\frac{x^4}{16}+\frac{1}{x^4}+\frac{1}{2}=\frac{1}{x^4}\left(\frac{x^8}{16}+1+\frac{x^4}{2}\right). $$ Let $y=x^4$, then the part in parentheses is $$ \frac{y^2}{16}+\frac{y}{2}+1=\left(\frac{y}{4}+1\right)^2. $$ Now, convert back to $x$'s and you'll need to multiply through by $\frac{1}{x^4}$. Remember that when bring the $\frac{1}{x^4}$ inside the square, you'll be multiplying by $\frac{1}{x^2}$.