$ \int\limits_{-\infty}^{+\infty} \frac{2}{(x-1)\cdot(x^2-6x+10)}\,\mathrm{d}\,x$ how to approach such types of problems as improper integral?

$$ \int\limits_{-\infty}^{+\infty} \frac{2}{(x-1)\cdot(x^2-6x+10)}\,\mathrm{d}\,x$$

is it also solve by complex theory integration?
 A: You could try partial fraction decomposition as suggested in the comments, but that doesn't immediately look like progress -- it will split into integrals that are even more improper, since $\int \frac{A}{x-1}\,dx$ diverges towards $\pm\infty$ in addition to around the pole at $x=1$.
Contour integration, on the other hand, looks promising:
For some large $R$, integrate along a contour from $-R$ to $1-\varepsilon$, then a small semicircle around the pole to $1+\varepsilon$, then straight on to $+R$, and a large semicircle back to $-R$ through the upper half-plane.
What we really want is the sum of the straight parts as $R\to\infty$ and $\varepsilon\to 0^+$. However, since the magnitude of the integrand falls away as $|x|^{-3}$, the contribution of the large semicircle goes to $0$ a $R\to\infty$. And the contribution of the small semicircle will tend to a purely imaginary value (why?) when $\varepsilon$ is small.
You already know the answer is real, so what you're looking for must be the real part of the integral along a contour that goes around the pole at $x=3+i$, which you can find easily using residues.
A: Your integral is$$\int_{-\infty}^\infty\left(\tfrac{2}{5(x-1)}-\tfrac{2(x-5)}{5(x^2-6x+10)}\right)dx=\left[\tfrac15\ln\tfrac{(x-1)^2}{x^2-6x+10}+\tfrac45\arctan(x-3)\right]_{-\infty}^\infty.$$As $x\to\pm\infty$, the numerator and denominator of $\tfrac{(x-1)^2}{x^2-6x+10}$ are both asymptotic to $x^2$, so the logarithm $\to\ln1=0$, and the integral is $\tfrac45(\tfrac{\pi}{2}-\tfrac{-\pi}{2})=\tfrac{4\pi}{5}$.
