<s>[INST] Integral of $\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}$ So we have to evaluate $\int\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}dx$. 
My work-

We can write the integrand as $\frac{(x+1)^2-2x}{(1-x)(1+x)\sqrt{1+x^4}}dx$. 
So we wish to deduce $\int\frac{(x+1)}{(1-x)\sqrt{1+x^4}}dx-\int\frac{2x}{(1-x^2)\sqrt{1+x^4}}dx$ 
So lets write it as $I_1+I_2$.

I tried and I can't evaluate $I_1$. 
For $I_2$, I took $x^2=t;\ 2xdx=dt$. 
The integral becomes $\int\frac{dt}{(1-t)\sqrt{1+t^2}}dt$. 
Okay, now we do $t=\tan\theta; \ dt=\sec^2\theta d\theta.$ 
$\int\frac{\sec\theta}{1-\tan\theta} d\theta$ 
Now what to do? Please guide me.?
 [/INST] If we replace $x$ by $\sqrt{u}$ we have:
$$ \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,dx=\int\frac{(1+u)}{2(1-u)\sqrt{u(1+u^2)}}\,du$$
Then, if we replace $\frac{1+u}{1-u}$ by $v$ we have:
$$ \int\frac{(1+u)}{2(1-u)\sqrt{u(1+u^2)}}\,du =\frac{1}{2\sqrt{2}}\int\frac{v}{\sqrt{v^4-1}}\,dv=\frac{1}{2\sqrt{2}}\,\log\left(v^2+\sqrt{v^4-1}\right)$$
so:

$$ \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,dx= \frac{1}{2\sqrt{2}}\log\left(\left(\frac{x^2+1}{x^2-1}\right)^2+\sqrt{\left(\frac{x^2+1}{x^2-1}\right)^4-1}\right).$$

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