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The integral to be solved is $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$. I tried substitution $t=x^8$ which got me to $\frac{1}{8} \int \sqrt{t + 2 + \frac{1}{t}} t^{-7/8} \,\mathrm{d}t$ and I'm stuck. Can you help?

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    $\begingroup$ What is $(x^4 +1/x^4)^2$? $\endgroup$ – JH vd Walt Jan 14 '18 at 15:53
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Note that$$x^8+2+x^{-8}=(x^4+x^{-4})^2.$$Can you take it from here?

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\begin{align} \int \sqrt{x^8 + 2 + x^{-8}} dx &= \int \sqrt{\left(x^4+\frac1{x^4}\right)^2} dx \\ &= \int \left|\underbrace{x^4+\frac1{x^4}}_{\ge0}\right| dx \\ &= \int \left(x^4+\frac1{x^4}\right) dx \\ &= \frac{x^5}5-\frac1{3x^3}+C \end{align}

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