Evaluation of $\int^{1}_{0}\frac{x^4}{1+x^8}dx$ 
How can I Integrate $\displaystyle \int^{1}_{0}\frac{x^4}{1+x^8}dx$?

I have searched in that forum and get the result for $\displaystyle \int^{\infty}_{-\infty}\frac{x^4}{1+x^8}dx$ or for $\displaystyle \int^{\infty}_{0}\frac{x^4}{1+x^8}dx$
Although i know the formula $\displaystyle \int^{\infty}_{0}\frac{x^{m-1}}{1+x^n}dx=\frac{\pi}{n}\cdot \csc\frac{\pi m}{n}$
But did not understand How can i use in my original Integration
Help me please Thanks

Note 
$$\int^{1}_{0}\frac{x^4}{1+x^8}dx
=\frac1{2\sqrt2}\int_0^1 \left(\frac{x^2}{x^4-\sqrt2x^2+1}
 - \frac{x^2}{x^4+\sqrt2x^2+1}\right)dx $$
where
$$\begin{align}
I(a)& =\int_0^1 \frac{x^2}{x^4+ax^2+1}dx \\
&=\frac12\int_0^1 \frac{x^2+1}{x^4+ax^2+1}dx -
\frac12\int_0^1 \frac{1-x^2}{x^4+ax^2+1}dx \\
&=\left(\frac1{2\sqrt{2+a}}\tan^{-1}\frac{x-\frac1x}{\sqrt{2+a}} 
 - \frac1{2\sqrt{2-a}}\coth^{-1}\frac{x+\frac1x}{\sqrt{2-a}} 
\right)_0^1 \\
&=\frac\pi{4\sqrt{2+a}}
 - \frac1{2\sqrt{2-a}}\coth^{-1}\frac{2}{\sqrt{2-a}} \\
\end{align}$$
Thus,
$$\begin{align}
&\int^{1}_{0}\frac{x^4}{1+x^8}dx
=\frac1{2\sqrt2}\left[I(-\sqrt2)-I(\sqrt2)\right]\\
=&\frac{\sqrt2}{8\sqrt{2-\sqrt2}}\left( \frac\pi2
+ \coth^{-1}\frac{2}{\sqrt{2-\sqrt2}} \right)
 -\frac{\sqrt2}{8\sqrt{2+\sqrt2}}\left( \frac\pi2+\coth^{-1}\frac{2}{\sqrt{2+\sqrt2}}\right) \\
\end{align}$$
