Tích phân $I=\int\limits_{0}^{1}{\frac{1}{\sqrt{{{x}^{2}}+x+1}}dx}$ bằng A. $\ln \frac{3+2\sqrt{3}}{3}.$ B. $\ln \frac{3-2\sqrt{3}}{3}.$ C. $\ln (9+6\sqrt{3}).$ D. $\ln \frac{3+\sqrt{3}}{3}.$
Đáp án đúng: A $t=x+\sqrt{{{x}^{2}}+x+1}=>\left\{ \begin{array}{l}dt=1+\frac{2x+1}{2\sqrt{{{x}^{2}}+x+1}}dx\\t=2{{x}^{2}}+x+1+2x\sqrt{{{x}^{2}}+x+1}\end{array} \right..$