Google
Câu hỏi: Cho $I=\int\limits_{1}^{2}{2x\sqrt{{{x}^{2}}-1}\text{d}x}$. Nếu đặt $u={{x}^{2}}-1$ thì khẳng định nào sau đây đúng?
A. $I=\dfrac{1}{2}\int\limits_{0}^{3}{\sqrt{u}\text{d}u}$.
B. $I=\int\limits_{1}^{2}{\sqrt{u}\text{d}u}$.
C. $I=\int\limits_{0}^{3}{\sqrt{u}\text{d}u}$.
D. $I=2\int\limits_{0}^{3}{\sqrt{u}\text{d}u}.$
A. $I=\dfrac{1}{2}\int\limits_{0}^{3}{\sqrt{u}\text{d}u}$.
B. $I=\int\limits_{1}^{2}{\sqrt{u}\text{d}u}$.
C. $I=\int\limits_{0}^{3}{\sqrt{u}\text{d}u}$.
D. $I=2\int\limits_{0}^{3}{\sqrt{u}\text{d}u}.$
Đặt $u={{x}^{2}}-1$, ta có $x=1\Rightarrow u=0; x=2\Rightarrow u=3$
Ta có: $u={{x}^{2}}-1\Rightarrow \text{d}u=2x\text{d}x\Rightarrow I=\int\limits_{0}^{3}{\sqrt{u}\text{d}u}$.
Ta có: $u={{x}^{2}}-1\Rightarrow \text{d}u=2x\text{d}x\Rightarrow I=\int\limits_{0}^{3}{\sqrt{u}\text{d}u}$.
Đáp án C.