Integral $\int\frac{dx}{x^5+1}$ Calculate $\displaystyle\int\dfrac{dx}{x^5+1}$
 A: One might as well explain how to derive the factorization in Troy Woo's answer. (This is by hand, not software.) Yes, he's right that any polynomial with real coefficients can be factorized into linear and quadratic factors with real coefficients; this fact can be derived from the fundamental theorem of algebra, by factorizing into linear factors $x - r_i$ where the $r_i$ range over all complex roots, and collecting factors that involve non-real roots into complex conjugate pairs in order to get at the real quadratic factors. 
Applying this to the polynomial $x^5 + 1$, let $\zeta$ be a complex fifth root of $-1$, say $\zeta = e^{\pi i/5}$. The other complex (non-real) fifth roots are $\bar{\zeta} = \zeta^{-1}$, $\zeta^3 = -\zeta^{-2}$, and $\bar{\zeta^3} = -\zeta^2$. Then 
$$x^5 + 1 = (x+1)(x - \zeta)(x - \zeta^{-1})(x + \zeta^2)(x + \zeta^{-2})$$ 
where we may observe 
$$0 = 1 - \zeta - \zeta^{-1} + \zeta^2 + \zeta^{-2}$$ 
by matching coefficients of $x^4$ on both sides. One of the conjugate pairs gives a real quadratic factor $(x - \zeta)(x - \zeta^{-1}) = x^2 - \alpha x + 1$. Here the real number $\alpha = \zeta + \zeta^{-1}$ can be computed as follows: 
$$\alpha^2 = (\zeta + \zeta^{-1})^2 = \zeta^2 + 2 + \zeta^{-2} = 2 + (-1 + \zeta + \zeta^{-1}) = 1 + \alpha$$ 
where the third equation is by the observation above. Solving for $\alpha$ gives either $\frac{1 + \sqrt{5}}{2}$ or $\frac{1 - \sqrt{5}}{2}$; actually it's the first if we start with the fifth root $\zeta = e^{\pi i/5}$. The other conjugate pair leads to $(x + \zeta^2)(x + \zeta^{-2}) = x^2 - \beta x + 1$ where $\beta$ is the other root $\frac{1 - \sqrt{5}}{2}$. This gives Troy Woo's answer. 
A: Hint:
$$
\begin{split}
x^5+1&=(1 + x)(x^4 - x^3 + x^2 - x + 1) \\
&=(1 + x)(x^2 + ((-1 + \sqrt 5) / 2)x + 1)(x^2 + ((-1 - \sqrt 5) / 2)x + 1) 
\end{split}
$$
A: $$x^5+1=(x+1)(x^4-x^3+x^2-x+1)\\=(x+1)\Big(x^2+\frac12(-1+\sqrt{5})x+1\Big)\Big(x^2-\frac12(1+\sqrt{5})+1\Big)$$
A: 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.
