Tích phân $I=\int\limits_{0}^{1}{{\frac{{xdx}}{{{{x}^{4}}+{{x}^{2}}+1}}}}$ bằng? A. $\frac{\pi }{6}.$ B. $\frac{\pi }{{\sqrt{3}}}.$ C. $\frac{\pi }{{6\sqrt{3}}}.$ D. $\frac{\pi }{{3\sqrt{3}}}.$
Đáp án đúng: C Đặt $t={{x}^{2}}=>I=\frac{1}{2}\int\limits_{0}^{1}{{\frac{{dt}}{{{{t}^{2}}+t+1}}}}=\frac{1}{2}\int\limits_{0}^{1}{{\frac{{dt}}{{{{{\left( {t+\frac{1}{2}} \right)}}^{2}}+{{{\left( {\frac{{\sqrt{3}}}{2}} \right)}}^{2}}}}=\frac{\pi }{{6\sqrt{3}}}.}}$