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Ta có
$\dfrac{1}{x\sqrt{x^2+x+1}}=\dfrac{1}{x}-\dfrac{1+\dfrac{1+2 x}{\sqrt{1+x+x^2}}}{{2+x+2 \sqrt{1+x+x^2}}}$
$\dfrac{1}{x\sqrt{x^2+x+1}}=(\ln x)'-\left (\ln(2+x+2 \sqrt{1+x+x^2}) \right )'$
Suy ra
$\int\limits_{1}^{+\infty }\dfrac{dx}{x\sqrt{x^2+x+1}}=\left[ {\ln x-\ln(2+x+2 \sqrt{1+x+x^2})} \right]_{1}^{+\infty }=\left[ {\ln\dfrac {x}{2+x+2 \sqrt{1+x+x^2}}} \right]_{1}^{+\infty }=-\ln(2\sqrt 3-3)$
Chú ý rằng
$\lim_{x \to +\infty }\dfrac {x}{2+x+2 \sqrt{1+x+x^2}}=\dfrac{1}{3}$
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