<s>[INST] How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods.  Then I learnt a very useful trick to evaluate the integral.
Noting that
$$I(1):= \int_{0}^{\infty} \frac{d x}{x^{4}-x^{2}+1} \stackrel{x \mapsto \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{2}}{x^{4}-x^{2}+1}
$$
Combining them yields
\begin{aligned}
I(1)&=\frac{1}{2} \int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\\&= \frac{1}{2}\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&= \frac{1}{2}\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1} \\
&= \frac{1}{2}\tan ^{-1}\left(x-\frac{1}{x}\right)_{0}^{\infty} \\
&= \frac{\pi}{2}
\end{aligned}
Later, I started to investigate the integrands with higher powers.
Similarly,
$$
I(2):= \int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{2}} \stackrel{x \mapsto \frac{1}{x}}{=}\int_{0}^{\infty} \frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}} d x
$$
By division, we decomposed $x^6$ and obtain $$
\frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}}=\frac{x^{2}+1}{x^{4}-x^{2}+1}-\frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}
$$$$
I(2)=\int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x-\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}dx
$$
We can now conclude that $$I(2)=I(1)=\frac{\pi}{2} $$
My Question:
How about the integral $$\displaystyle I_{n}=\int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}$$ for any integer $n\geq 3$?
 [/INST] With CAS and help of MellinTransfrom:
$$\int_0^{\infty } \frac{1}{\left(x^4-x^2+1\right)^n} \, dx=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \mathcal{M}_a\left[\frac{1}{\left(x^4-a x^2+1\right)^n}\right](s) \,
   dx\right](1)=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \frac{(-1)^{-s} x^{-2 s} \left(1+x^4\right)^{-n+s} \Gamma (n-s) \Gamma (s)}{\Gamma (n)} \,
   dx\right](1)=\\\mathcal{M}_s^{-1}\left[\frac{(-1)^{-s} \Gamma \left(\frac{1}{4}-\frac{s}{2}\right) \Gamma \left(-\frac{1}{4}+n-\frac{s}{2}\right) \Gamma (s)}{4 \Gamma
   (n)}\right](1)=\\\frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(-\frac{1}{4}+n\right) \, _2F_1\left(\frac{1}{4},-\frac{1}{4}+n;\frac{1}{2};\frac{1}{4}\right)}{4 \Gamma
   (n)}+\frac{\Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{1}{4}+n\right) \, _2F_1\left(\frac{3}{4},\frac{1}{4}+n;\frac{3}{2};\frac{1}{4}\right)}{4 \Gamma (n)}$$
</s>