$ \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. 

This is not a final answer but it is too long for a comment.
Beside the elegant solution given by Dr.MV using the residue theorem, there is something I found interesting in the problem. $$\frac{1+x^6}{1-x^2+x^4-x^6+x^8}=(1+x^2+x^6+x^8)\sum_{k=0}^{\infty}(-1)^k x^{10k}$$ So $$I=\int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,dx=\sum_{k=0}^{\infty}\int_{0}^1 (-1)^k(1+x^2+x^6+x^8)x^{10k}\,dx$$ which leads to $$I=\sum_{k=0}^{\infty}(-1)^k \left(\frac{1}{10 k+1}+\frac{1}{10 k+3}+\frac{1}{10 k+7}+\frac{1}{10
   k+9}\right)$$ and $I$ has a very nice and simple expression.
