Evaluate $\int \frac 1{x^{12}+1} \, dx$ 
Evaluate $\displaystyle \int \frac 1{x^{12}+1} \, dx$

I tried writing this in partial fractions.
$$\int \frac 1{x^{12}+1} \, dx=\int \frac{1}{[(x^6+1)+\sqrt{2}x^3][(x^6+1)-\sqrt{2}x^3]} \, dx$$
So I did:
\begin{align}
\frac{1}{[(x^6+1)+\sqrt{2}x^3][(x^6+1)-\sqrt{2}x^3]} &= \frac{Ax^5+Bx^4+Cx^3+Dx^2+Ex+F}{[(x^6+1)-\sqrt{2}x^3]}\\&{}\qquad+\frac{A_1x^5+B_1x^4+C_1x^3+D_1x^2+E_1x+F_1}{[(x^6+1)+\sqrt{2}x^3]}\end{align}
After messy work, I found $A=A_1=B=B_1=D=D_1=E=E_1=0$ and
$$C=-\frac 1{2\sqrt{2}},\qquad C_1=\frac 1{2\sqrt{2}},\qquad F=F_1=\frac 12$$
I now get the integral
$$\int  \frac{-\frac 1{2\sqrt{2}}x^3+\frac 12}{(x^6+1)-\sqrt{2}x^3}\, dx + \int \frac{\frac 1{2\sqrt{2}}x^3+\frac 12}{(x^6+1)+\sqrt{2}x^3} \, dx$$
But how can I go from here? I am stuck.
 A: Answer 1 - use computer algebra
At the level of just getting some answer for a relatively simple rational function, I would suggest to use some symbolic calculation tools like Mathematica, Wolfram Alpha.
In particular for the question you ask above, Mathematicia/Wolfram alpha gives you an explicit (real-coefficient) anti-derivative:
http://www.wolframalpha.com/input/?i=primitive+for+1%2F%28x%5E12+%2B+1%29
(Sorry about just providing the link - the formula is rather long or I'd type it in here.  My point though is that - at this point int time - there is software readily available, even online, that will get you an answer pretty much right away.)  
Answer above to factor into primitive roots of unity and integrate will work nicely but may be difficult to get a real coefficient solution out of it?
If you're interested in actually going through the actual calculation by hand, some of the answers address that so i won't go into it.  Seems like it will take quite a bit of work.  But having an answer may help you get there (look at the pieces, take derivatives).
Answer 2 - exploit that the roots are roots of unity
Sorry, should have thought of this right away.
The roots will be simple roots of unity coming in conjugate pairs, which you can exploit to factor the polynomial into a product of quadratic factors (with real coefficients) using the fact that (x-c)(x-cbar) = x - 2 Re(c) + |c|^2 (where cbar is the conjugate of c):
e.g.
x^8 - x^4 + 1 =
= product of the prime 12th roots of -1 (1st, 5th, 7th, 11th roots of -1 and their conjugates)
=(x^2 + 2 cos(pi/12)x +1)(x^2 + 2 cos(11pi/12)x +1)(x^2 + 2 cos(5pi/12)x +1)(x^2 + 2 cos(7pi/12)x +1)
so 
x^12 + 1 
= (x^4+1)(x^8 - x^4 + 1)
= above * (x^2 + 2cos(pi/4)x + 1)*(x^2 + 2cos(3pi/4)x + 1)
which is a complete (real!) factorization of the denominator.  You can verify the factorization with computer algebra.  That should allow you to proceed with partial fractions(?) - still with quite a bit of pain but at least you have the denominators.
A: Decompose and arrange the integrand as follows
\begin{align}\frac{6}{1+x^{12}}
& =\frac2{x^4+1}-\frac{\sqrt3x^2-2}{x^4-\sqrt3x^2+1} 
 +\frac{\sqrt3x^2+2}{x^4+\sqrt3x^2+1} \\
 & =\frac{(1+x^2)+(1-x^2)}{x^4+1}+\frac{2+\sqrt3}2\left(\frac{1+x^2}{x^4+\sqrt3x^2+1}+\frac{1-x^2}{x^4-\sqrt3x^2+1} \right)\\
& \hspace{30mm}+\frac{2-\sqrt3}2\left(\frac{1+x^2}{x^4-\sqrt3x^2+1}+\frac{1-x^2}{x^4+\sqrt3x^2+1} \right)\\
\end{align}
Then
\begin{align}
\int \frac{dx}{1+x^{12}}
& =\frac16J(0)
+\frac{2+\sqrt3}{12}J(\sqrt3) +\frac{2-\sqrt3}{12}J(-\sqrt3) +C\\
\end{align}
where
\begin{align}
J(a)&= \int \left(\frac{1+x^2}{x^4+ax^2+1}+ \frac{1-x^2}{x^4-ax^2+1}\right)dx \\
&=\int \frac{d(x-\frac1x)}{(x-\frac1x)^2+(2+a)}+\int \frac{d(x+\frac1x)}{(2+a)-(x+\frac1x)^2}  \\
&=\frac1{\sqrt{2+a}}\left( \tan^{-1}\frac{x^2-1}{x\sqrt{2+a}} +\coth^{-1}\frac{x^2+1}{x\sqrt{2+a}} \right)\\
\end{align}
A: Hint: solve $x^{12}=-1$ in the complex plane, then use partial fractions.
Note that $(-1)^{1/12}=\cos{\left(\frac{2k\pi+\pi}{12}\right)}+i\sin{\left(\frac{2k\pi+\pi}{12}\right)}$, with $k=0,1,2,\cdots,11$
Now the roots are: 
$$
        \begin{matrix}
        x_1=\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}i, & x_2=\frac{\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{6}-\sqrt{2}}{4}i  \\
        x_3=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i, & x_4=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i  \\
        x_5=\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4}i, &x_6=\frac{\sqrt{6}-\sqrt{2}}{4}-\frac{\sqrt{6}+\sqrt{2}}{4}i \\
        x_7=\frac{-\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4}i, & x_8=\frac{-\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{6}+\sqrt{2}}{4}i  \\
        x_9=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i, & x_{10}=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i   \\
        x_{11}=\frac{-\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}i, & x_{12}=\frac{-\sqrt{6}-\sqrt{2}}{4}-\frac{\sqrt{6}-\sqrt{2}}{4}i  
        \end{matrix}
$$
by the factor theorem:
$$x^{12}+1=(x-x_1)(x-x_2)\cdots(x-x_{11})(x-x_{12})$$
further it is known that $(x+a+bi)(x+a-bi)=x^2+2ax+a^2+b^2$,
Then $$x^{12}+1=(x^2+m_1x+n_1)(x^2+m_2x+n_2)\cdots(x^2+m_6x+n_6)$$
Now apply partial fractions
A: The denominator of the fraction has twelve complex roots which are of the form $\omega^k$, where $\omega$ is a twelfth root of minus one.
By decomposition in simple fractions,
$$\frac1{x^{12}+1}=\frac1{12}\left(\frac{\omega^1}{x-\omega^1}+\frac{\omega^3}{x-\omega^3}+\cdots\frac{\omega^{23}}{x-\omega^{23}}\right).$$
Now, as
$$\frac1{x-\omega}=\frac1{x-r-is}=\frac{x-r+is}{x^2-2rx+r^2+s^2}=\frac{y+is}{y^2+s^2}.$$
we can integrate to obtain
$$\log(x-\omega)=\log\sqrt{y^2+s^2}+i\arctan\frac ys=\log\sqrt{(x-r)^2+s^2}+i\arctan\frac {x-r}s$$
and you can sum the twelve contributions. By symmetry, they
can be grouped in pairs and you get a sum of six logarithms and six arc-tangents.
A: $x^{12}+1=(x^4+1)(x^8-x^4+1)$ and $x^4+1=(x^2+x\sqrt{2}+1)(x^2-\sqrt{2}x+1)$
further we have $x^8-x^4+1=- \left( \sqrt {2}{x}^{3}+{x}^{4}+x\sqrt {2}+{x}^{2}+1 \right) 
 \left( \sqrt {2}{x}^{3}-{x}^{4}+x\sqrt {2}-{x}^{2}-1 \right) 
$
