<s>[INST] Evaluating $\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$ I tried substitution $x=\dfrac{1}{t}$, then $z=t^2$, tried rationalizing the denominator but haven't been able to pull it off. I can't think of any trig substitution either. I prefer an intuitive approach rather than an "plucked-out-of-thin-air" counter-intuitive substitution.
 [/INST] Here is an elementary method without using hyperbolic functions:
$$\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$$
$$=\int \frac{1}{x^5(\frac{1}{x^4}+1)\sqrt{\sqrt{\frac{1}{x^4}+1}-1}}dx$$
Now substitute $t=\frac{1}{x^4}+1$, required integral becomes,
$$-\frac{1}{4}\int\frac{1}{t\sqrt{\sqrt {t}-1}}dt$$
Now substitute $y=\sqrt{\sqrt {t}-1}$, to get,
$$t=(y^2+1)^2$$
$$\implies dt=4y(y^2+1)dy$$
$$\implies \text{Required integral}=-\frac{1}{4}\int\frac{4y(y^2+1)}{y(y^2+1)^2}dy$$
$$=-\tan^{-1}y+C$$
$$=-\tan^{-1}\left(\frac{\sqrt{\sqrt{x^4+1}-x^2}}{x}\right)+C$$
$$=\tan^{-1}\left(\frac{x}{\sqrt{\sqrt{x^4+1}-x^2}}\right)+C'$$
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