<s>[INST] How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$ $$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$ 
This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?
 [/INST] It might not be elliptic after all... (unless I have made some mistake)
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
Let $\displaystyle u = x -\frac{1}{x}$.
Then $\displaystyle du = (1 + \frac{1}{x^2})dx$.
Now $$\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}} = -\frac{x^2(1 + 1/x^2)}{x(x-1/x)\sqrt{x^2(x^2 + 1/x^2)}} = -\frac{1 + 1/x^2}{(x-1/x)\sqrt{(x - 1/x)^2 + 2}}$$
Thus
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
$$= -\int \ \frac{\mathrm{du}}{u \sqrt{u^2 + 2}}$$
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