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.

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, $$t=(y^2+1)^2$$ $$\implies dt=4y(y^2+1)dy$$ $$\implies \text{Required integral}=-\frac{1}{4}\int\frac{4y(y^2+1)}{y(y^2+1)^2}dy$$ $$=-\tan^{-1}y+C$$ $$=-\tan^{-1}\left(\frac{\sqrt{\sqrt{x^4+1}-x^2}}{x}\right)+C$$ $$=\tan^{-1}\left(\frac{x}{\sqrt{\sqrt{x^4+1}-x^2}}\right)+C'$$