$\displaystyle \int\limits_{1}^{2}{{\frac{x}{{1+\sqrt{{x-1}}}}dx}}$ bằng : A. $\frac{{11}}{3}+\ln 2$ B. $\frac{{11}}{3}-\ln 2$ C. $\frac{{11}}{3}+4\ln 2$ D. $\frac{{11}}{3}-4\ln 2$
Đáp án đúng: D Đặt $\sqrt{x-1}=t\Rightarrow 2tdt=dx.$ Ta có$I=2\int\limits_{0}^{1}{t.\frac{{{t}^{2}}+1}{t+1}dt}=2\int\limits_{0}^{1}{\left( {{t}^{2}}-t+2-\frac{2}{1+t} \right)dt}=\frac{11}{3}-4\ln 2.$