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$\int\limits_{\sqrt{2}}^{+\infty }\dfrac{x dx}{(x^2+1)^3}=\dfrac{
1}{2}\int\limits_{\sqrt{2}}^{+\infty }\dfrac{ d(x^2+1)}{(x^2+1)^3}=\left[ {-\dfrac{
1}{4}\dfrac{1}{(x^2+1)^2}} \right]_{\sqrt{2}}^{+\infty }=\dfrac{1}{36}$
Chú ý rằng
$\lim_{x \to +\infty}\dfrac{1}{(x^2+1)^2}=0$
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