Decomposition into partial fractions to compute an integral

Question

I'm having problems with:

$$\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}dx$$

I was thinking: $\frac{x^4+1}{x^6+1}$ is an even function and the interval $(-\infty,\infty)$ is symmetric about 0, we could write the integral like:

$$2\int_{0}^{\infty}\frac{(x^4+1)}{x^6+1}dx$$

And for $\frac{x^4+1}{x^6+1}$, I will use the partial fractions. I will write $x^6+1$ like a sum of cubes $x^6+1=(x^2)^3+1^3$ and use the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and in the end we will have $$x^6+1=(x^2+1)(x^4-x^2+1)$$ But now I'm stuck and do not know how to decompose $x^4-x^2+1$. I was thinking about $(x^2-1)x^2+1$ or $(x^2-\frac{1}{2})^2+\frac{3}{4}$ but I need a form for that like $-(-1+\sqrt3 x-x^2) (1+\sqrt3 x+x^2)$...

A little help here?

Answer 1

HINT:

As $y^3+1=(y+1)(y^2-y+1),$

$$\dfrac{x^4+1}{x^6+1}=\dfrac{x^4-x^2+1}{x^6+1}+\dfrac{x^2}{x^6+1}=\dfrac1{x^2+1}+\dfrac{x^2}{x^6+1}$$

Set $x^3=u$ for the second part

Answer 2

Another way,

Solution :

\begin{align} \int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx&=2\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx\\[7pt] &=2\int_{0}^{\infty}\frac{x^4\,\mathrm dx}{x^6+1}+2\int_{0}^{\infty}\frac{\mathrm dx}{x^6+1}\tag{$\color{red}{❤}$}\\[7pt] &=\frac{\pi}{3\sin\left(\frac{5\pi}{6}\right)}+\frac{\pi}{3\sin\left(\frac{\pi}{6}\right)}\\[7pt] &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{4\pi}{3}}} \end{align}

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Explanation :

$(\color{red}{❤})\;$ $\displaystyle\int_0^\infty\frac{x^{\large n-1}}{x^m+1}\ \,\mathrm dx=\frac{\pi}{m\sin\left(\frac{n\pi}{m}\right)}\,$ for $\,0<n<m$.

Answer 3

$$\begin{eqnarray*}\int_{\mathbb{R}}\frac{x^4+1}{x^6+1}\,dx&=&2\int_{\mathbb{R}^+}\frac{x^4+1}{x^6+1}\,dx=2\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx+2\int_{1}^{+\infty}\frac{x^4+1}{x^6+1}\,dx\\&=&4\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx=4\int_{0}^{1}\sum_{k\geq 0}(-1)^k\left(x^{6k}+x^{6k+4}\right)\,dx\\&=&4\sum_{k\geq 0}(-1)^k\left(\frac{1}{6k+1}+\frac{1}{6k+5}\right)=4\sum_{k\geq 0}\frac{\chi(k)}{k}\end{eqnarray*}$$ where $\chi(k)$ is a Dirichlet character $\!\!\pmod{12}$, that equals one if $k\in\{1,5\}\pmod{12}$, minus one if $k\in\{7,11\}\pmod{12}$ and zero otherwise. We know in advance that: $$ \sum_{k\geq 0}\frac{(-1)^k}{2k+1}=\arctan 1=\frac{\pi}{4},$$ $$ \sum_{k\geq 0}\frac{(-1)^k}{6k+3}=\frac{1}{3}\,\arctan 1=\frac{\pi}{12},$$ hence by summing these series we get $\sum_{k\geq 0}(-1)^k\left(\frac{1}{6k+1}+\frac{1}{6k+5}\right)=\frac{\pi}{3}$, so:

$$ \int_{\mathbb{R}}\frac{x^4+1}{x^6+1}\,dx = \color{red}{\frac{4\pi}{3}}.$$