Use the Residue theorem and its application to compute the integral $$\int_{-\infty}^{\infty} \frac{x^2}{x^4-4x^2+5} dx. $$
I am not sure how to approach this question. Can anyone use the complex variable theory to help me solving the problem please?
Thank you very much. 
 A: HINT:  The roots of $x^4 - 4x^2 + 5$ in the upper half-plane are $\sqrt{2 + i}$ and $- \sqrt{2 - i}$ (to factor, think about $x^4 - 4x^2 + 5$ as a quadratic in the variable $x^2$).  Then think about integrating over a semicircular contour $\gamma = [- R, R] \cup S$, where
$$ S = \{ R e^{it} : 0 \leq t \leq \pi \}. $$
Let $R \to \infty$.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\int_{-\infty}^{\infty}{x^{2} \over x^{4} - 4x^{2} + 5}\,\dd x
=
\int_{-\infty}^{\infty}
{x^{2}
 \over
 \bracks{x^{2} - \pars{2 - \ic}}\bracks{x^{2} - \pars{2 + \ic}}}\,\dd x
\\[3mm]&=
\int_{-\infty}^{\infty}
{x^{2}
 \over
 \pars{x^{2} - \root{5}\expo{-\ic\phi}}\pars{x^{2} - \root{5}\expo{\ic\phi}}}\,\dd x
 \quad\mbox{where}\quad\phi = \arctan\pars{1/2}.
\end{align}

\begin{align}
&\int_{-\infty}^{\infty}{x^{2} \over x^{4} - 4x^{2} + 5}\,\dd x
\\[3mm]&=
\int_{-\infty}^{\infty}
{x^{2}
 \over
 \pars{x - 5^{1/4}\expo{-\ic\phi/2}}
 \bracks{x - \pars{\color{#ff0000}{-5^{1/4}\expo{-\ic\phi/2}}}}
 \pars{x - \color{#ff0000}{5^{1/4}\expo{\ic\phi/2}}}
 \bracks{x - \pars{-5^{1/4}\expo{\ic\phi/2}}}} 
\,\dd x
\end{align}
The $\color{#ff0000}{\large\mbox{red}}$ terms are the poles in the upper complex half plane. So, can you take it from here ?.

