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
$$\int{1\over ax^2+bx+c}dx={2\over \sqrt{4ac-b^2}}\tan^{-1}{2ax+b\over \sqrt{4ac-b^2}}$$

Elaborating user @Dr. MV's answer, we have
\begin{equation}
\int_0^\infty\frac{1}{a^2x^4+bx^2+c^2}\ dx=\frac{c\pi}{2a\sqrt{b+2ac}}
\end{equation}
Putting $a=1$, $b=a$, and $c^2=b$, then
\begin{equation}
I(a,b)=\int_0^\infty\frac{1}{x^4+ax^2+b}\ dx=\frac{\pi}{2}\sqrt{\frac{b}{a+2\sqrt{b}}}
\end{equation}
Hence
\begin{equation}
\frac{\partial}{\partial a}I(a,b)=\int_0^\infty\frac{x^2}{\left(x^4+ax^2+b\right)^2}\ dx=-\frac{\pi}{2b}\sqrt{\left(\!\frac{b}{a+2\sqrt{b}}\!\right)^3}
\end{equation}
and
\begin{equation}
\frac{\partial}{\partial b}I(a,b)=\int_0^\infty\frac{1}{\left(x^4+ax^2+b\right)^2}\ dx=\frac{\pi(a+\sqrt{b})}{2\sqrt{b\left(a+2\sqrt{b}\right)^3}}
\end{equation}
