Evaluation of $\int \frac{\sqrt{1+x^4}}{1-x^4}dx$ Evaluation of $\displaystyle \int \frac{\sqrt{1+x^4}}{1-x^4}dx$
$\bf{My\; Try::}$ Given $\displaystyle \int\frac{\sqrt{1+x^4}}{1-x^4}dx\;,$ Then  We can write the above Integral as $$\displaystyle \int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}dx = \frac{1}{2}\int\left\{\frac{1}{1-x^2}+\frac{1}{1+x^2}\right\}\cdot \frac{(1+x^4)}{\sqrt{1+x^4}}dx$$
So Integral is $$\displaystyle = \frac{1}{2}\int\frac{1+x^4}{(1-x^2)\sqrt{1+x^4}}dx+\frac{1}{2}\int\frac{1+x^4}{(1+x^2)\sqrt{1+x^4}}dx$$
Now Let $$\displaystyle I = \int\frac{1+x^4}{(1-x^2)\sqrt{1+x^4}}dx$$ and $$\displaystyle J = \int\frac{1+x^4}{(1+x^2)\sqrt{1+x^4}}dx$$
Now How can I evaluate $I$ and $J\;,$ plz help me
Thanks

Thanks  Claude Leibovici and Lucian, Using Lucian Hint.
$\bf{Another \; Try::}$ Let $\displaystyle \mathcal{I} = \int \frac{\sqrt{1+x^4}}{1-x^4}dx\;,$ Now Let $x^2=\tan \phi\;,$
Then $\displaystyle 2xdx = \sec^2 \phi d\phi \Rightarrow dx = \frac{\sec^2 \phi}{2\sqrt{\tan \phi}}d\phi$.
So $$\displaystyle \mathcal{I} = \int\frac{\sec \phi}{1-\tan^2 \phi}\times \frac{\sec^2 \phi}{2\sqrt{\tan \phi}}d\phi = \frac{1}{2}\int \frac{\sec \phi}{\cos 2\phi}\times \frac{1}{\sqrt{\tan \phi}}d\phi = \frac{\sqrt{2}}{2}\int\frac{1}{\cos 2\phi\cdot \sqrt{\sin 2\phi}}d\phi$$
So $$\displaystyle \mathcal{I} = \frac{1}{2}\int\frac{\cos 2\phi}{\left(1-\sin^2 2\phi\right)\cdot \sqrt{\sin 2\phi}}d\phi$$
Now Let $\displaystyle \sin 2\phi = t^2\;,$ Then $\displaystyle \cos 2\phi d\phi = 2tdt$.
So $$\displaystyle \mathcal{I} = \frac{1}{2}\int\frac{1}{1-t^4}dt = \frac{1}{4}\int \left(\frac{1}{1-t^2}+\frac{1}{1+t^2}\right)dt = \frac{1}{4}\cdot \frac{1}{2}\ln \left|\frac{1+t}{1-t}\right|+\frac{1}{4}\tan^{-1}(t)+\mathcal{C}$$
So $$\displaystyle \mathcal{I} = \frac{1}{4}\cdot \frac{1}{2}\ln \left|\frac{1+\sqrt{\sin 2\phi}}{1-\sqrt{\sin 2\phi}}\right|+\frac{1}{4}\tan^{-1}(\sqrt{\sin 2\phi})+\mathcal{C}$$
Where $\phi = \tan^{-1}(x^2).$
