Tích phân $I=\int\limits_{0}^{1}{\sqrt{\frac{3-x}{1+x}}dx}$ bằng A. $\frac{\pi }{3}-\sqrt{2}+2.$ B. $\frac{\pi }{3}-\sqrt{3}+2.$ C. $\frac{\pi }{3}-\sqrt{3}+1.$ D. $\frac{\pi }{4}-\sqrt{3}+2.$
Đáp án đúng: B Đặt $t=\sqrt{\frac{3-x}{x+1}}.$ $\Rightarrow 2tdt=-\frac{4}{x+1}dx.$ Nên ta có $\begin{array}{l}I=\int\limits_{\sqrt{3}}^{1}{t.\frac{2t}{-{{t}^{2}}-1}dt}=\int\limits_{1}^{\sqrt{3}}{\frac{2{{t}^{2}}}{{{t}^{2}}+1}dt}\\=\left. \left( 2t-2\ln \left( 1+{{t}^{2}} \right) \right) \right|_{1}^{\sqrt{3}}=\frac{\pi }{3}-\sqrt{3}+2.\end{array}$