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.?

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.
