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.

Split the integral into two integrals, such that you have only one "bad" point. Now:
$$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx= \int_{-\infty}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx + \int_{0}^{\infty}{{x^2-3x+2}\over {x^4+10x^2+9}}dx$$
Now deal with the integrals separately:
$$\int_{-\infty}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to -\infty} \int_{t}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to -\infty} \frac{1}{48}\left(9\ln\left(\frac{x^2+9}{x^2+1}\right) + 6\arctan(x) + 14\arctan(\frac x3)\right) \Biggr|_t^0 = \frac{9}{48}\ln(9) + \frac{10\pi}{48}$$
$$\int_{0}^{\infty} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to \infty} \int_{0}^{t} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to \infty} \frac{1}{48}\left(9\ln\left(\frac{x^2+9}{x^2+1}\right) + 6\arctan(x) + 14\arctan(\frac x3)\right) \Biggr|_0^t = -\frac{9}{48}\ln(9) + \frac{10\pi}{48}$$
Summing them you will get:
$$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \frac{9}{48}\ln(9) + \frac{10\pi}{48} -\frac{9}{48}\ln(9) + \frac{10\pi}{48} = \frac{5\pi}{12}$$
NOTE: I skipped the calculation of the integrals, but it can be easily done by partial fraction decomposition and it will be reduced to something more familiar.
