Evaluation of $\displaystyle \int\frac{1}{1+x^6}dx$   $\bf{My\; Try::}$Let $$I = \int\frac{1}{1+x^6}dx = \int\frac{1}{(1+x^2)(x^4-x^2+1)}dx$$  Using Partial fraction , above Integral is very lengthy, Can we solve it without using   partial fraction or any other way, Help me  Thanks
I came across following problem     Evaluate $$\int\frac{1}{1+x^6} \,dx$$   When I asked my teacher for hint he said first evaluate     $$\int\frac{1}{1+x^4} \,dx$$   I've tried to factorize $1+x^6$ as  $$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$ and then writing  $$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$ $$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$  However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$ But I can't see how it helps  I've also tried to reverse engineer the   And I need to have terms similar to $$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?  Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?
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