# How to find $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$?

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?

• What is $(x^4 +1/x^4)^2$? – JH vd Walt Jan 14 '18 at 15:53

Note that$$x^8+2+x^{-8}=(x^4+x^{-4})^2.$$Can you take it from here?