Proving that $\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\frac{\pi}{3}$, where $\{.\}$ is positive fractional part

Question

Here, $\{-3.4\}=0.6$. The said integral can be solved using $\{z\}+\{-z\}=1$, if $z$ is a non-zero real number; after using the property that $$\int_{-a}^{a} f(x) dx= \int_{0}^{a} [ f(x)+f(-x)] dx$$ So here $$I=\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\int_{0}^{1} \frac{[\{x^3\}+\{-x^3\}](x^4+1)}{(x^6+1)} dx =\int_{0}^{1} \frac{(x^4+1)}{(x^6+1)} dx.$$ $$\implies I= \int_{0}^{1} \frac{(1+x^2)^2-2x^2}{(x^6+1)}dx=\int_{0}^{1}\frac{(1+x^2) dx}{x^4-x^2+1}-\int_{0}^{1} \frac{2x^2 dx}{x^6+1}=I_1-I_2.$$ In $I_1$, divide up and down by $x^2$ and use $x-1/x=u$, then $$I_1=\int_{-\infty}^{0} \frac{du}{1+u^2}=\frac{\pi}{2}$$ Next, use $x^3=v$ $$I_2=\int_{0}^{1} \frac{2x^2 dx}{x^6+1}=\frac{2}{3} \int_{0}^{1} \frac{dv}{1+v^2}=\frac{\pi}{6}$$ Finally $$I=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$$

It will be interesting to see other approaches/methods of proving this integral.

Answer 1

$$I=\int\dfrac{x^4+1}{x^6+1}\,dx=\int\dfrac{x^4+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\,dx=\int\left(\dfrac{x^2+1}{3\left(x^4-x^2+1\right)}+\dfrac{2}{3\left(x^2+1\right)}\right)dx$$ $$\int\dfrac{x^2+1}{x^4-x^2+1}\,dx=\int\dfrac{x^2+1}{\left(x^2-\sqrt{3}x+1\right)\left(x^2+\sqrt{3}x+1\right)}\,dx=\int\left(\dfrac{1}{2\left(x^2+\sqrt{3}x+1\right)}+\dfrac{1}{2\left(x^2-\sqrt{3}x+1\right)}\right)dx$$ Complete the squares and finish to get $$I=\dfrac{\arctan\left(2x+\sqrt{3}\right)+\arctan\left(2x-\sqrt{3}\right)+2\arctan\left(x\right)}{3}$$ Combine the arctangents to end with $$I=\frac{1}{3} \left(\tan ^{-1}\left(\frac{x}{1-x^2}\right)+2 \tan ^{-1}(x)\right)$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x} = \int_{0}^{1}{1 + x^{4} - x^{6} - x^{10} \over 1 - x^{12}}\,\dd x \\[5mm] = &\ {1 \over 12}\int_{0}^{1}{x^{-11/12} + x^{-7/12} - x^{-5/12} - x^{-1/12} \over 1 - x}\,\dd x \\[5mm] = &\ {1 \over 12}\bracks{-\Psi\pars{1 \over 12} - \Psi\pars{5 \over 12} + \Psi\pars{7 \over 12} + \Psi\pars{11 \over 12}} \\[5mm] = &\ {1 \over 12}\braces{\bracks{% \Psi\pars{7 \over 12} - \Psi\pars{5 \over 12}} + \bracks{\Psi\pars{11 \over 12} - \Psi\pars{1 \over 12}}} \\[5mm] = &\ {1 \over 12}\bracks{\pi\cot\pars{5\pi \over 12} + \pi\cot\pars{\pi \over 12}} = {\pi \over 12}{\sin\pars{5\pi/12 + \pi/12} \over \sin\pars{5\pi/12}\sin\pars{\pi/12}} \\[5mm] = &\ {\pi \over 12}{1 \over \bracks{\vphantom{\Large A}\cos\pars{5\pi/12 - \pi/12} - \cos\pars{5\pi/12 + \pi/12}}/\, 2} \\[5mm] = &\ {\pi \over 6}{1 \over \cos\pars{\pi/3} - \cos\pars{\pi/2}} = {\pi \over 6}{1 \over 1/2 - 0} = \bbox[10px,#ffd,border:1px groove navy]{\large{\pi \over 3}} \\ & \end{align}