$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$ How do we compute this integral ?
$$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
I have tried partial fraction but it is quite hard to factorize the denominator. Any help is appreciated. 

Let $I$ be the integral given by 
$$I= \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
Now, make the substitution $x=u^{-1}$. Then $dx=-u^{-2}du$ and the limits of integration over $u$ extend from $\infty$ to $1$.  Thus, we may write
$$
\begin{align}
I & = \int_{\infty}^1 \frac{1+u^{-6}}{1-u^{-2}+u^{-4}-u^{-6}+u^{-8}}\left(-u^{-2}\right)du \\
& = \int_1^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du 
\end{align}
$$
whereby 
$$I=\frac12 \int_0^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du$$
Next, we exploit the even symmetry of the integrand, which reveals that 
$$I=\frac14 \int_{-\infty}^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du$$
Using the Residue Theorem along with Jordan's Lemma, the integral $I$ is given by
$$I=2\pi i \left(\frac14\right)\sum_{i=1}^4 Res_i \left(\frac{1+z^{6}}{1-z^{2}+z^{4}-z^{6}+z^{8}}\right)$$
where the summation is over the four residues of the integrand in the upper-half plane.  
All that remains is to find the four complex roots $z_i$ of the denominator that lie in the upper-half plane.  These are $z_1=e^{\frac{\pi}{10}}$, $z_2=e^{\frac{3\pi}{10}}$, $z_3=e^{\frac{7\pi}{10}}$, and $z_4=e^{\frac{9\pi}{10}}$.
To find the residue of $z_i$, take the limit
$$\lim_{z \to z_i} (z-z_i) \left(\frac{1+z^{6}}{1-z^{2}+z^{4}-z^{6}+z^{8}}\right)=\frac{1+z_i^{6}}{-2z_i+4z_i^{3}-6z_i^{5}+8z_i^{7}}$$
