<s>[INST] Evaluate $\int \frac{\sqrt{1+x^8} dx}{x^{13}}$ 
Evaluate: $\displaystyle\int \frac{\sqrt{1+x^8}}{x^{13}}dx$

My attempt:
I have tried substituting $1+x^8=z^2$ but that did not work. I also tried writing $x^{13}$ in the denominator as $x^{16}.x^{-3}$ hoping that it would bring the integrand into some form but that too did not work.
 [/INST] The substitution $x=\tan^{1/4}t$ gives $$\int\frac{\sqrt{1+x^8}}{x^{13}}dx=\int\frac{1}{4}\sin^{-4}t\cos tdt=-\frac{1}{12}\sin^{-3}t+C=-\frac{1}{12}\bigg(\frac{x^8}{1+x^8}\bigg)^{-3/2}+C.$$
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