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?
 A: What a surprise! 
Surfing the net, I found an almost same question on
"hard integral"
$$
\displaystyle \int \frac{x^2 - 1}{(x^2 + 1) \sqrt{x^4 + 1}} \, dx
$$
from June 19, 2008.
A: 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
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
$$= -\int \ \frac{\mathrm{du}}{u \sqrt{u^2 + 2}}$$
A: Somewhat inspired by Moron's wonderful answer, I decided to see if a trigonometric solution would do the job.
Making the substitution $x=\cot\left(\frac{\theta}{2}\right)$, $\mathrm dx=\frac{\mathrm d\theta}{\cos\;\theta-1}$, we have
$$\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\cot^4\frac{\theta}{2}}}$$
$$=\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\left(\frac{1+\cos\;\theta}{1-\cos\;\theta}\right)^2}}$$
$$=\frac1{\sqrt{2}}\int \frac{\mathrm d\theta}{\cos\;\theta\sqrt{1+\cos^2\theta}}$$
which integrates to
$$\frac1{\sqrt{2}}\tanh^{-1}\frac{\sin\;\theta}{\sqrt{1+\cos^2\theta}}$$
Undoing the substitution, we get
$$\frac1{\sqrt{2}}\tanh^{-1}\left(x\sqrt{\frac{2}{x^4+1}}\right)$$
and it is easy to verify that the derivative of this last expression gives the original integrand.
A: 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
$$
\frac{1}{2\sqrt{2}}\ln\left(\frac{(1+x^2)^2+2\sqrt{2}x\sqrt{1+x^4}}{(1-x^2)^2}\right)+C.
$$
