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Câu hỏi: Cho tích phân $I=\int\limits_{0}^{3}{\dfrac{x}{1+\sqrt{x+1}}}\text{d}x$ nếu đặt $t=\sqrt{x+1}$ thì $I$ là
A. $I=\int\limits_{1}^{2}{\left( 2{{t}^{2}}-2t \right)}\text{dt}$.
B. $I=\int\limits_{1}^{2}{\left( 2{{t}^{2}}-t \right)}\text{dt}$.
C. $I=\int\limits_{1}^{2}{\left( 2{{t}^{2}}+2t \right)}\text{dt}$.
D. $I=\int\limits_{1}^{2}{\left( {{t}^{2}}-2t \right)}\text{dt}$.
A. $I=\int\limits_{1}^{2}{\left( 2{{t}^{2}}-2t \right)}\text{dt}$.
B. $I=\int\limits_{1}^{2}{\left( 2{{t}^{2}}-t \right)}\text{dt}$.
C. $I=\int\limits_{1}^{2}{\left( 2{{t}^{2}}+2t \right)}\text{dt}$.
D. $I=\int\limits_{1}^{2}{\left( {{t}^{2}}-2t \right)}\text{dt}$.
đặt $t=\sqrt{x+1}\Rightarrow {{t}^{2}}=x+1\Leftrightarrow 2t\text{d}t=\text{d}x$.
$x=0\Rightarrow t=1$ ;
$x=3\Rightarrow t=2$.
$I=\int\limits_{1}^{2}{\dfrac{{{t}^{2}}-1}{1+t}}\text{2}t\text{d}t\text{=}\int\limits_{1}^{2}{\left( \text{2}{{t}^{2}}-2t \right)}\text{d}t$.
$x=0\Rightarrow t=1$ ;
$x=3\Rightarrow t=2$.
$I=\int\limits_{1}^{2}{\dfrac{{{t}^{2}}-1}{1+t}}\text{2}t\text{d}t\text{=}\int\limits_{1}^{2}{\left( \text{2}{{t}^{2}}-2t \right)}\text{d}t$.
Đáp án A.