Câu hỏi 38

\(I=\int\limits_{0}^{1}{\frac{dt}{{{t}^{2}}+t+1}}=\int\limits_{0}^{1}{\frac{dt}{{{\left( t+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}}\)

Đặt \(x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan x\Leftrightarrow dt=\frac{\sqrt{3}}{2}\left( 1+{{\tan }^{2}}x \right)dx\)

Đổi cận \(\left\{ \begin{array}{l}t = 0 \Rightarrow x = \frac{\pi }{6}\\t = 1 \Rightarrow x = \frac{\pi }{3}\end{array} \right.\), khi đó ta có \(I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\frac{\sqrt{3}}{2}\left( 1+{{\tan }^{2}}x \right)dx}{\frac{3}{4}\left( 1+{{\tan }^{2}}x \right)}}=\left. \frac{2}{\sqrt{3}}t \right|_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{2}{\sqrt{3}}\frac{\pi }{6}=\frac{\pi \sqrt{3}}{9}\)

Chọn B.