Partial Fractions of $\frac{1}{x^6+1}$

Question

I am trying to solve the following integral : $$\int \frac{1}{1+x^6}dx$$

I do not want the reader to evaluate the integral, but rather the partial fractions of the integrand : $\frac{1}{1+x^6}$.

This is my workings for factoring the integrand : $$\begin{align}x^6+ 1 &= (x^2+1)(x^4-x^2+1)\\ &= (x^2+1)(x^4+2x^2+1-3x^2)\\ &=(x^2+1)((x^2+1)^2-(\sqrt{3}x)^2)\\ &=(x^2+1)(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)\end{align}$$

Therefore : $$\int\frac{1}{x^6+1}dx=\int\frac{a_1x+a_2}{x^2+1}dx+\int\frac{a_3x+a_4}{x^2-\sqrt{3}x+1}dx+\int\frac{a_5x+a_6}{x^2+\sqrt{3}x+1}dx$$

From here, I am not sure how to proceed to solve for $a_n$.

Answer 1

You could work this out from there by bringing the terms back to a common fraction.

$$ \frac{a_1 x + a_2}{x^2+1} + \frac{a_3 x + a_4}{x^2 - \sqrt{3} x + 1} + \frac{a_5 x + a_6}{x^2 + \sqrt{3} x + 1} = \frac{1}{x^6+1} $$

for all $x$ if and only if

$$ (a_1 x + a_2)(x^2 - \sqrt{3} x + 1)(x^2 + \sqrt{3} x + 1) + (a_3 x + a_4)(x^2 + 1)(x^2 + \sqrt{3} x + 1) + (a_5 x + a_6)(x^2 + 1)(x^2 - \sqrt{3} x + 1) = 1 $$

Multiplying it all out and collecting like terms makes the left side a degree 5 polynomial, and the coefficients must be $0, 0, 0, 0, 0, 1$. So that's six linear equations in six unknowns.

But that's going to be a huge mess!

To keep this more manageable, let's try computing numerators with the denominator only partly factored, not all the way into irreducibles over the reals. Since $x^6+1 = (x^2+1)(x^4-x^2+1)$, we know there are real $a_0, b_0, b_2$ so that

$$ \frac{a_0}{x^2+1} + \frac{b_2 x^2 + b_0}{x^4-x^2+1} = \frac{1}{x^6+1} $$

If you're wondering why we don't have an $x$ term above $x^2+1$ and an $x^3$ and an $x$ term above $x^4-x^2+1$, it's because I've treated $x^2$ rather than $x$ as the "free" variable here. That is, we can find a representation of $1/(y^3+1)$ in terms of $1/(y+1)$ and $1/(y^2-y+1)$ which is valid for all real $y$, then substitute $y=x^2$ to get the wanted formula for $1/(x^6+1)$; it will have only even powers of $x$.

Multiplying out gives

$$ a_0(x^4 - x^2 + 1) + (b_2 x^2 + b_0)(x^2+1) = 1 $$ $$ (a_0+b_2) x^4 + (b_0+b_2-a_0) x^2 + (a_0+b_0) = 1 $$ $$ a_0 + b_2 = 0 \qquad b_0+b_2-a_0 = 0 \qquad a_0+b_0 = 1 $$

The unique solution to these is $a_0 = \frac{1}{3}$, $b_2 = -\frac{1}{3}$, $b_0 = \frac{2}{3}$. So

$$ \frac{1}{x^6+1} = \frac{1}{3}\left[\frac{1}{x^2+1} + \frac{2-x^2}{x^4 - x^2 + 1}\right] $$

Now for the next factor split, let's look for coefficients so that

$$ \frac{c_1 x+c_0}{x^2 - \sqrt{3} x + 1} + \frac{d_1 x+d_0}{x^2 + \sqrt{3} x + 1} = \frac{2-x^2}{x^4-x^2+1} $$ $$ (c_1 x+c_0)(x^2 + \sqrt{3}x + 1) + (d_1 x+d_0)(x^2 - \sqrt{3} x + 1) = 2-x^2 $$ $$ (c_1+d_1)x^3 + (c_1 \sqrt{3} + c_0 - d_1 \sqrt{3} + d_0)x^2 + (c_1 + c_0 \sqrt{3} + d_1 - d_0 \sqrt{3}) x + (c_0+d_0) = -x^2 + 2 $$ $$ c_1+d_1=0 \qquad c_1 \sqrt{3} + c_0 - d_1 \sqrt{3} + d_0 = -1 \qquad c_1 + c_0 \sqrt{3} + d_1 - d_0 \sqrt{3} = 0 \qquad c_0+d_0=2 $$

I get the unique solution as $c_0=1$, $c_1=-\sqrt{3}/2$, $d_0=1$, $d_1=\sqrt{3}/2$. So

$$ \frac{2-x^2}{x^4-x^2+1} = \frac{1 - \frac{\sqrt{3}}{2} x}{x^2-\sqrt{3}x+1} + \frac{1 + \frac{\sqrt{3}}{2} x}{x^2+\sqrt{3}x+1} $$

Finally,

$$ \frac{1}{x^6+1} = \frac{\frac{1}{3}}{x^2+1} + \frac{\frac{1}{3} - \frac{1}{2\sqrt{3}} x}{x^2-\sqrt{3}x+1} + \frac{\frac{1}{3} + \frac{1}{2\sqrt{3}} x}{x^2+\sqrt{3}x+1} $$

Answer 2

In traditional solution, there are totally 6 unknowns with 6 equations to be solved. The work is rather complicated and tedious. However, we can simplify the work by resolving, instead of the original, $$ \frac{3}{x^{3}+1}=\frac{1}{x+1}+\frac{2-x}{x^{2}-x+1} $$ *refer to the footnote for detailed proof.

Then replacing $x$ by $x^2$ yields $$ \frac{3}{x^{6}+1}=\frac{1}{x^{2}+1}+\frac{2-x^{2}}{x^{4}-x^{2}+1}, $$ from which we immediately get $$ a_{1}=0 \text{ and }a_{2}=\frac{1}{3}. $$

Now we are going to further resolve the second one into 2 partial fractions.

$$ \begin{aligned}&\text{Let } \quad \frac{2-x^{2}}{x^{4}-x^{2}+1} \equiv \frac{a_{3} x+a_{4}}{x^{2}-\sqrt{3} x+1}+\frac{a_{5} x+a_{6}}{x^{2}+\sqrt{3} x+1} \\ &\text{then }\quad 2-x^{2} \equiv\left(a_{3} x+a_{4}\right)\left(x^{2}+\sqrt{3} x+1\right)+\left(a_{5} x+a_{6}\right)\left(x^{2}-\sqrt{3} x+1\right) & \end{aligned} $$

Comparing the coefficient of $x^3$and the constant yields $$ \left\{\begin{array}{l} a_{3}+a_{5}=0 \cdots(1) \\ a_{4}+a_{6}=2 \cdots(2) \end{array}\right. $$

Putting $x=i$ yields

$$ \begin{aligned} 2-i^{2}=&\left(a_{3} i+a_{4}\right)(\sqrt{3} i)+\left(a_{5} i+a_{6}\right)(-\sqrt{3} i) \\ 3=&-a_{3} \sqrt{3}+a_{4} \sqrt{3} i+a_{5} \sqrt{3}-a_{6} \sqrt{3} i \end{aligned} $$

Comparing the real parts yields $$ a_{5}-a_{3}=\sqrt{3} \cdots(3) $$

Similarly, putting $x= -i$ and comparing the imaginary parts yields $$ -a_{4}+a_{6}=0 \cdots(4) $$ (1) +(3) yields $$ a_{3}=-\frac{\sqrt{3}}{2} \text { and } a_{5}=\frac{\sqrt{3}}{2}$$

(2) -(4) yields $$ a_{4}=1 \text { and } a_{6}=1 $$

Now we can conclude that $$ \frac{1}{x^{6}+1}=\frac{1}{3\left(x^{2}+1\right)}+\frac{2-\sqrt{3} x}{6\left(x^{2}-\sqrt{3} x+1\right)}+\frac{2+\sqrt{3} x}{6\left(x^{2}+\sqrt{3} x+1\right)} $$

By the way, the integrand doesn’t need to be resolved into 3 fractions. We can integrate $$ \int \frac{3}{x^{6}+1} d x=\int \frac{d x}{x^{2}+1}+\int \frac{2-x^{2}}{x^{4}-x^{2}+1} d x $$

By my post,

\begin{aligned} \int \frac{1}{x^{6}+1} d x=& \frac{1}{3} \tan ^{-1} x+\frac{1}{6} \tan ^{-1}\left(x-\frac{1}{x}\right) +\frac{\sqrt{3}}{12} \ln \left|\frac{x^{2}+\sqrt{3} x+1}{x^{2}-\sqrt{3} x+1}\right|+C \end{aligned}

Footnote:

\begin{array}{l} \text {Let } \displaystyle \frac{3}{x^{3}+1} \equiv \frac{A}{x+1}+\frac{f(x)}{x^{2}-x+1} ,\text{ then }\\ \qquad 3 \equiv A\left(x^{2}-x+1\right)+f(x)(x+1). \\ \text{Putting }{ x=-1 } \text{ yields }1=A(1+1+1) \Rightarrow A=1 \\ \text {Hence } \displaystyle f(x)=\frac{3-\left(x^{2}-x+1\right)}{x+1}=2-x \end{array}

Answer 3

Here is a trick that makes calculations faster. We have, $$\frac1{x^6+1}=\frac{a_1 x + a_2}{x^2+1} + \frac{a_3 x + a_4}{x^2 - \sqrt{3} x + 1} + \frac{a_5 x + a_6}{x^2 + \sqrt{3} x + 1} $$ Note that $f(x)=\dfrac{1}{x^6+1}$ is an even function, Hence on the right side of above equation, changing $x$ to $-x$ doesn't change the expression. Hence $a_1=0\;, \;a_3=-a_5\;$ and $\;a_4=a_6$. Therefore number of coefficients will be reduced to three. $$\frac{1}{x^6+1}=\frac{A}{x^2+1}+\frac{Bx+C}{x^2-\sqrt3x+1}+\frac{-Bx+C}{x^2+\sqrt3x+1}$$ From here you can expand RHS or alternatively plug in three small values of $x$ (like $x=0,1,-1$) in each side and solve system of equations in three variables $A,B,C$.