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.?

$\bf{My\; Solution::}$ Given $$\displaystyle I = \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}dx = \int\frac{1+x^2}{x^2\cdot \left(\frac{1}{x}-x\right)\sqrt{\left(\frac{1}{x}-x\right)^2+\left(\sqrt{2}\right)^2}}dx$$
So $$\displaystyle I = \int\frac{\left(\frac{1}{x^2}+1\right)}{\left(\frac{1}{x}-x\right)\sqrt{\left(\frac{1}{x}-x\right)^2+\left(\sqrt{2}\right)^2}}dx$$
Now Put $\displaystyle \left(\frac{1}{x}-x\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt$
So Integral $$\displaystyle I = \int\frac{1}{t\sqrt{t^2+\left(\sqrt{2}\right)^2}}dt$$
Now Put $t = \sqrt{2}\tan \theta \;,$ Then $dx = \sqrt{2} \sec^2 \theta d\theta$
So $$\displaystyle I = \int\frac{\sqrt{2}\sec^2 \theta }{2\tan \theta \cdot \sec \theta }d\theta =\frac{1}{\sqrt{2}}\int \csc \theta d\theta  = -\frac{1}{\sqrt{2}}\ln \left|\csc \theta + \cot \theta \right|+\mathcal{C}$$
So $$\displaystyle I = -\frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{2}+\sqrt{t^2+2}}{t}\right|+\mathcal{C}\;,$$ Where $$\displaystyle \left(\frac{1}{x}-x\right) = t.$$
