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.?
 A: 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:

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

A: My method : 
Am guiding you in different way which is too easy to understand 
Make it a try ... 
First step : remove $x^2$ from numerator and $x$ from first factor and $x^2$ from under square root function
Then it becomes $\int\frac{1+\frac1{x^2}}{(\frac1{x} -x)\sqrt{x^2+\frac{1}{x^2}}}$ 
Next 
Why don't you take $\frac1{x}-x=t; -(\frac{1}{x^2}+1)dx=dt)$ 
 Then it changes 
$\int \frac{-dt}{t \sqrt{t^2+2}}$
Again take  $t^2 +2=u^2$ 
  $2t dt =2udu$
$t dt= u du$
Now use these then problem becomes  $-\int \frac{du}{u^2-2}$
Using standard results we get 
$\frac{-1}{2\sqrt{2}}.In|{\frac{u-\sqrt2}{u+\sqrt2}}|$
Change $u$ to $t$ and change $t$ to $x$ that's ur answer 
May i think you like it.
A: $\bf{My\; Solution::}$ 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 $$\displaystyle \left(\frac{1}{x}-x\right) = t.$$
