Show that $\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}$ Show that $$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}.$$
Any solutions or hints are greatly appreciated. 
I know I can rewrite the integral as  $$\int_{-\infty}^\infty {(x-1)(x-2)\over {(x^2+1)(x^2+9)}}dx.$$ but I'm not sure how to proceed.

Well, $\int_{-\infty}^{+\infty}\frac{-3x}{x^4+10x^2+9}\,dx = 0$, hence the problem boils down to computing:
$$ \int_{-\infty}^{+\infty}\frac{(x^2+2)}{(x^2+1)(x^2+9)}\,dx=\frac{1}{8}\int_{-\infty}^{+\infty}\frac{1}{x^2+1}\,dx+\frac{7}{8}\int_{-\infty}^{+\infty}\frac{1}{x^2+9} $$
that is trivially equal to $\left(\frac{1}{8}+\frac{7}{24}\right)\pi = \color{red}{\large\frac{5\pi}{12}}$.

Footnote: how to find the coefficients $\frac{1}{8}$ and $\frac{7}{8}$ very fast. We know that for some $A,B$
$$ g(z)=\frac{z+2}{(z+1)(z+9)}=\frac{A}{z+1}+\frac{B}{z+9} $$
must hold. On the other hand, $A=\lim_{z\to -1}g(z)(z+1)$ as well as $B=\lim_{z\to -9}g(z)(z+9)$, so:
$$ A = \lim_{z\to -1}\frac{z+2}{z+9},\qquad B=\lim_{z\to -9}\frac{z+2}{z+1}.$$
