$\int\limits_0^\infty {x^4 \over (x^4-x^2+1)^4}\ dx$ I want to calculate
$$\int\limits_0^\infty \frac{x^4}{(x^4-x^2+1)^4}dx$$
I have searched with keywords "\frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results
 A: For $b > 0$, define $I_b(a)$ by
$$ I_b(a)
= \int_{0}^{\infty} \frac{x^{a}}{((x - x^{-1})^2 + 1)^b} \, \mathrm{d}x
= \int_{0}^{\infty} \frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} \, \mathrm{d}x. $$
This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x \mapsto 1/x$. Then
\begin{align*}
\int_{0}^{\infty} \frac{x^4}{(x^4 - x^2 + 1)^4} \, \mathrm{d}x
&= I_4(-4)
 = I_4(2) \\
&= I_4(2) - I_4(0) + I_4(-2) \\
&= \int_{0}^{\infty} \frac{1}{((x - x^{-1})^2 + 1)^3} \, \mathrm{d}x
\end{align*}
So, by the Glasser's master theorem,
\begin{align*}
\int_{0}^{\infty} \frac{x^4}{(x^4 - x^2 + 1)^4} \, \mathrm{d}x
&= \int_{0}^{\infty} \frac{1}{(u^2 + 1)^3} \, \mathrm{d}u \\
&= \int_{0}^{\frac{\pi}{2}} \cos^4\theta \, \mathrm{d}\theta \tag{$(u=\tan\theta)$} \\
&= \frac{3\pi}{16}
 \approx 0.58904862254808623221 \cdots.
\end{align*}
A: Here is another way to get to the same point as what @Sangchul Lee gives.
Let
$$I = \int_0^\infty \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx.$$
Then
$$I = \int_0^1 \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx + \int_1^\infty \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx.$$
Enforcing a substitution of $x \mapsto 1/x$ in the right most integral leads to
\begin{align}
I &= \int_0^1 \frac{x^4 (1 + x^6)}{(x^4 - x^2 + 1)^4} \, dx\\
&= \int_0^1 \frac{x^4 (1 + x^2)(x^4 - x^2 + 1)}{(x^4 - x^2 + 1)^4} \, dx\\
&= \int_0^1 \frac{x^4 (1 + x^2)}{(x^4 - x^2 + 1)^3} \, dx\\
&= \int_0^1 \frac{1 + 1/x^2}{\left [\left (x - \frac{1}{x} \right )^2 + 1 \right ]^3} \, dx.
\end{align}
On setting $-u = x - 1/x$, $-du = (1 + 1/x^2) \, dx$ one has
\begin{align}
I &= \int_0^\infty \frac{du}{(u^2 + 1)^3}\\
&= \int_0^{\frac{\pi}{2}} \cos^4 \theta \, d\theta\\
&= \int_0^{\frac{\pi}{2}} \left (\frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta + \frac{3}{8} \right ) \, d\theta\\
&= \frac{3\pi}{16},
\end{align}
as expected. 
