Tích phân $I=2\int\limits_{0}^{1}{\frac{{{x}^{2}}}{(x+1)\sqrt{x+1}}dx}$ bằng A. $\frac{16-13\sqrt{2}}{4}.$ B. $\frac{16-13\sqrt{2}}{3}.$ C. $\frac{16-11\sqrt{2}}{4}.$ D. $\frac{16-11\sqrt{2}}{3}.$
Đáp án đúng: D Đặt $t=\sqrt{x+1}=>dt=\frac{1}{2\sqrt{x+1}}dx.$ $\Rightarrow 2tdt=dx.$ Khi đó $\begin{array}{l}I=2\int\limits_{1}^{\sqrt{2}}{\frac{{{\left( {{t}^{2}}-1 \right)}^{2}}2tdt}{t.{{t}^{2}}}}=4\int\limits_{1}^{\sqrt{2}}{\left( {{t}^{2}}-2+\frac{1}{{{t}^{2}}} \right)dt}\\=\left. 4\left( \frac{{{t}^{3}}}{3}-2t-\frac{1}{t} \right) \right|_{1}^{\sqrt{2}}=\frac{16-11\sqrt{2}}{3}.\end{array}$