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Ta có
$\dfrac{1}{(x^2+1)^2}=\dfrac{1}{2}.\dfrac{1-x^2}{(x^2+1)^2}+\dfrac{1}{2}.\dfrac{1}{x^2+1}$
$\dfrac{1}{(x^2+1)^2}=\dfrac{1}{2}.\left (\dfrac{x}{x^2+1} \right )'+\dfrac{1}{2}.\left ( \arctan x \right )'$
Suy ra
$\int\limits_{0}^{\infty }\dfrac{dx}{(x^2+1)^2}=\dfrac{1}{2}\left[ {\dfrac{x}{x^2+1}+\arctan x} \right]_{0}^{\infty }=\dfrac{\pi}{4}$
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