<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] Moron's and J.M.'s solutions are nice.
Hopefully this solution is simpler.
Without loss of generality we may assume that $1\gt x\gt 0$.
Put $x:=\sqrt{y}$, $1\gt y\gt 0$. Then we obtain
$$
\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx=\int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy.
$$
Introduce the new variable 
$$
t:=\frac{1+y}{1-y},\qquad 1\lt t \lt \infty.
$$
Then we have
$$
y=\frac{-1+t}{1+t},
$$
$$
\mathrm dy=\frac{2}{(1+t)^2}\,\mathrm dt.
$$
Substituting back we obtain
$$
\int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy=\int{\frac{t}{2\sqrt{1+\left(\dfrac{-1+t}{1+t} \right)^2}\sqrt{\dfrac{-1+t}{1+t}}}\frac{2}{(1+t)^2}\,\mathrm dt}
$$
$$
=\frac{1}{\sqrt{2}}\int{\frac{t}{\sqrt{t^4-1}}}\mathrm dt
$$
$$
=\frac{1}{2\sqrt{2}}\ln(t^2+\sqrt{t^4-1})+C.
$$
Putting back everything we obtain
$$
\frac{1}{2\sqrt{2}}\ln\left(\frac{(1+x^2)^2+2\sqrt{2}x\sqrt{1+x^4}}{(1-x^2)^2}\right)+C.
$$
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