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

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\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}
