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How is this integral simplified as shown?

\begin{align} S &=2\pi \int^3_\frac{1}{2} \Bigg(\frac{x^3}{6} + \frac{1}{2x}\Bigg)\sqrt{1+\Bigg(\frac{x^2}{2} - \frac{1}{2x^2}\Bigg)^2} \mathop{\mathrm{d}x}\\ &= 2\pi\int^3_\frac{1}{2} \Bigg(\frac{x^5}{12}+\frac{x}{3} + \frac{1}{4x^3}\Bigg)\mathop{\mathrm{d}x} \end{align}

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    $\begingroup$ Since the simplification has nothing to do with the fact that you are integrating, or indeed with anything but rational-function algebra, why the calculus, integration, and definite-integrals tags? $\endgroup$
    – LSpice
    Jun 5, 2016 at 17:05

3 Answers 3

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Note that $$\left(\frac{x^2}{2}-\frac{1}{2x^2}\right)^2=\left(\frac{x^2}{2}\right)^2-2\cdot\frac{x^2}{2}\cdot\frac{1}{2x^2}+\left(\frac{1}{2x^2}\right)^2=\left(\frac{x^2}{2}\right)^2-\frac{1}{2}+\left(\frac{1}{2x^2}\right)^2.$$ So when you add $1$, the middle term becomes $+\frac{1}{2}$. Thus $$1+\left(\frac{x^2}{2}-\frac{1}{2x^2}\right)^2=\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)^2.$$

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$1+(\frac{x^2}{2}-\frac{1}{2x^2})=(\frac{x^2}{2}+\frac{1}{2x^2})^2$. Now you can do it

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Hint: Expand the term under the square root. It is a square, namely $(x^8 + 2x^4 + 1)/(4x^4)$. Do you see why ?

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