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$I=\int\limits_{-\infty }^{+\infty }\dfrac{dx}{x^2+4x+9}=\int\limits_{-\infty }^{+\infty }\dfrac{d(x+2)}{(x+2)^2+5}=\left[ {\dfrac{1}{\sqrt 5}\arctan\dfrac{x+2}{\sqrt 5}} \right]_{-\infty }^{+\infty }=\dfrac{\pi}{\sqrt 5}$
Trong đó đã sử dụng
$\arctan(+\infty)=\dfrac{\pi}{2},\arctan(-\infty)=\dfrac{-\pi}{2}$
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