Integral $\int\frac{dx}{x^5+1}$ Calculate $\displaystyle\int\dfrac{dx}{x^5+1}$

If you're not all about complex numbers, you could think of it this way.  It should be easy to see (if you know either polynomial long division or synthetic division; if not ask me) that $$x^5+1=(x+1)(x^4-x^3+x^2-x+1).$$
Now we want to think about how to factor the remaining part $x^4-x^3+x^2-x+1$.  Immediately upon looking at this it may not be clear what to do.  But for me, I see the $x^4$ term and think that the whole thing should factor as $x^2$ in one factor times $x^2$ in another.  I will suppose that is true and write
$$x^4-x^3+x^2-x+1=(x^2+Cx+1)(x^2+Dx+1)$$
The reason I have both those factors end in a 1 is because the left side of the equation has to end in 1 and that's the only way I could think to do it.  Now if you FOIL out the right side you get
$$x^4+(D+C)x^3+(2+CD)x^2+(D+C)x+1.$$
Compare the coefficients of the left side of the equation with those of the right side now and you should get
$$C+D=1\ \ \ and\ \ \ 2+CD=1$$
Solve for C and D and you will have the factorization.  I guess from there you would use partial fractions to complete the integral.
