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Câu hỏi: Xét $\int\limits_{0}^{1}{\left( x-1 \right){{e}^{{{x}^{2}}-2x+3}}\text{d}x}$, nếu đặt $u={{x}^{2}}-2x+3$ thì $\int\limits_{0}^{1}{\left( x-1 \right){{e}^{{{x}^{2}}-2x+3}}\text{d}x}$ bằng
A. $-\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
B. $\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
C. $-\dfrac{1}{2}\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
D. $\dfrac{1}{2}\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
A. $-\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
B. $\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
C. $-\dfrac{1}{2}\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
D. $\dfrac{1}{2}\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
Đặt $u={{x}^{2}}-2x+3\Rightarrow \text{d}u=2\left( x-1 \right)\text{d}x\Rightarrow \left( x-1 \right)\text{d}x=\dfrac{1}{2}\text{d}u$.
Đổi cận: $x=0\Rightarrow u=3$ ; $x=1\Rightarrow u=2$.
Ta có $\int\limits_{0}^{1}{\left( x-1 \right){{e}^{{{x}^{2}}-2x+3}}\text{d}x}=\int\limits_{3}^{2}{{{e}^{u}}\dfrac{1}{2}\text{d}u}=-\dfrac{1}{2}\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
Đổi cận: $x=0\Rightarrow u=3$ ; $x=1\Rightarrow u=2$.
Ta có $\int\limits_{0}^{1}{\left( x-1 \right){{e}^{{{x}^{2}}-2x+3}}\text{d}x}=\int\limits_{3}^{2}{{{e}^{u}}\dfrac{1}{2}\text{d}u}=-\dfrac{1}{2}\int\limits_{2}^{3}{{{e}^{u}}\text{d}u}$.
Đáp án C.