I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that
$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$
Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$
Let see (substitution of $y=x^2$)
$$\int_{-\infty}^{\infty}{x\over (x^4-x^2+1)^2}dx={1\over 2}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$
Substituion of $y=x^3$
$$\int_{-\infty}^{\infty}{x^3\over (x^4-x^2+1)^2}dx={1\over 4}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$
As for $\int_{-\infty}^{\infty}{x^2\over (x^4-x^2+1)^2}dx$ and $\int_{-\infty}^{\infty}{x^4\over (x^4-x^2+1)^2}dx$ are difficult to find a suitable substitution. This is the point where I am shrugged with to find a suitable substitution To lead me to a particular standard integral. Need some help, thank.
standard integral of the form
$$\int{1\over (ax^2+bx+c)^2}dx={2ax+b\over (4ac-b^2)(ax^2+bx+c)}+{2a\over 4ac-b^2}\int{1\over ax^2+bx+c}dx$$
And
$$\int{1\over ax^2+bx+c}dx={2\over \sqrt{4ac-b^2}}\tan^{-1}{2ax+b\over \sqrt{4ac-b^2}}$$

Evaluating the integral of interest can be reduced to evaluating the integral 
$$I(a,b)=\int_0^\infty \frac{1}{x^4+bx^2+a}  \,dx\tag 1$$
To see this, we exploit first odd symmetry to write the integral of interest as
$$\begin{align}
\int_{-\infty}^\infty \frac{(x^2-x+\pi)^2}{(x^4-x^2+1)^2}\,dx&=2\int_0^\infty \frac{x^4+(2\pi +1)x^2+\pi^2}{(x^4-x^2+1)^2}\,dx \tag 2
\end{align}$$
Enforcing the substitution $x\to 1/x$ in $(2)$ reveals
$$\begin{align}
\int_0^\infty \frac{x^4}{(x^4-x^2+1)^2}\,dx&=\int_0^\infty \frac{x^2}{(x^4-x^2+1)^2}\,dx\\\\
&=-\left.\frac{\partial I(a,b)}{\partial b}\right|_{(a,b)=(1,-1)} \tag 3
\end{align}$$
Moreover, we have
$$\int_0^\infty \frac{1}{(x^4-x^2+1)^2}\,dx=-\left.\frac{\partial I(a,b)}{\partial a}\right|_{(a,b)=(1,-1)} \tag 4$$
Using $(3)$ and $(4)$ in $(2)$ yields
$$\int_{-\infty}^\infty \frac{(x^2-x+\pi)^2}{(x^4-x^2+1)^2}\,dx=-4(\pi +1)\left.\frac{\partial I(a,b)}{\partial b}\right|_{(a,b)=(1,-1)}-2\pi^2\left.\frac{\partial I(a,b)}{\partial a}\right|_{(a,b)=(1,-1)}$$
where $I(a,b)$ as given by $(1)$ can be found using partial fraction expansion or contour integration, for examples.
