<s>[INST] 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
$$\int{1\over ax^2+bx+c}dx={2\over \sqrt{4ac-b^2}}\tan^{-1}{2ax+b\over \sqrt{4ac-b^2}}$$
 [/INST] First note that because of
$$
\int_{\mathbb{R}}dx\frac{x^4}{(x^4-x^2+1)^2}=\int_{\mathbb{R}}dx\frac{x^4\overbrace{-x^2+1+(x^2-1)}^{=0}}{(x^4-x^2+1)^2}=\\
\int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{x^2-1}{(x^4-x^2+1)^2}=\\
\int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{-1}{(x^4-x^2+1)^2}+\color{blue}{\underbrace{\int_{\mathbb{R}}dx\frac{x^2}{(x^4-x^2+1)^2}}_{J}}
$$
we only need to calulate $\color{\blue}{J}$ because everthing else is covered by your standard formulas. 
Now let's define
$$
I(a)=\int_{\mathbb{R}}dx\frac{1}{(x^4-a x^2+1)}
$$
from which it follows that
$$
\color{blue}{J}=I'(1)
$$
so it can also derived using ur standard identities ($'$ denotes a derivative w.r.t. $a$)
Edit:
Note that the integrals with numerators containing odd powers of $x$ vanish due to symmetry!
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