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I am learning about how to calculate the length of a path with integration. The equation is:

$\sqrt{1+\Big (\frac{dy}{dx} \Big)^2} $

So I have to integrate it between $a$ and $b$. In my book I have an example, I understand it but don't know how he solved this:

$\sqrt{\frac{1}{4} \times (x^4 + 2 + \frac{1}{x^4})}$

for which the result after eliminating the square root is:

$1/2 \times (x^2 + \frac{1}{x^2})$

I know that $\sqrt{1/4} = 1/2$ but I can't understand how the rest is solved.

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    @Andres : "Equation" is not the right word; the expression you have there is not an equation. Also, "solve" is not the right word. One solve problems; one solves equations; one _evaluates_ or _finds_ or _simplifies_ expressions.2012-06-21

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This is a special case of the familiar identity $a^2+2ab+b^2=(a+b)^2,$ with $a=x^2$ and $b=\frac{1}{x^2}$. That makes the "middle" term $2ab$ equal to $2$.

The easiest way to see things is probably to work backwards, and expand $\left(x^2+\frac{1}{x^2}\right)^2.$

Remark: In most cases, if we take a function $y$, and calculate $\sqrt{1+\left(\frac{dy}{dx}\right)^2}$, we get something ugly that cannot be integrated in terms of elementary functions. So arclength problems often involve fairly artificial functions $y$ for which $\sqrt{1+\left(\frac{dy}{dx}\right)^2}$ magically happens to simplify.